Bí quyết ôn luyện thi đại học đạt điểm tối đa vật lý tập 1 lê văn vinh part 2

h Phan xa cua song tren vat can t y do Thuc nghiem cho thay, khi phan xa tren vat can t u do bien dang cua day khong bj dao chieu.. Cac diem thupc cimg mpt bo song khoang giiia hai nut

Phuong trinh song tai mpt M la: u ^ = 2acos

Phuong trinh song tai mot O la:

C a u 5 : Hai nguon ke't hop Si,S2 each nhau mpt khoang 50mm tren mat nuoc

phat ra hai song ke't hop c6 phuong trinhuj = U 2 = 2cos2007it(mm) Van

toe truyen song tren mat nuoc la 0,8 m/s Diem gan nhat dao dpng ciing pha

voi nguon tren duong trung true eua S1S2 each nguon Si bao nhieu:

Giai nhanh bai nay n h u sau:

Diem M dao dong ciing pha voi nguon Si nen phai each Si mpt doan d = kX

Nhin tu hinh ve ta tha'y:

d = k^>S,0^8k>25:^k>3,125:^k„i„ = 4 => d^,„ = 4X = 4.8 = 32mm Chgn B

;au 6: Tren mat mot cha't long, c6 hai nguon song ke't hpp Oi, O2 each nhau

/ = 24em, d a o dong theo ciing mpt phuong voi phuong trinh

"oi = " o 2 = Aeoswt (t tinh bang s A tinh bang mm) Khoang each ngSn nhat

tu trung diem O eua O1O2 den cac diem nam tren duong trung true eua O1O2 dao dong ciing pha voi O b3ng q = 9cm So diem dao dong voi bien dp bang O tren doan O1O2 la:

A 18 B 16 C 20 D 14

^hdn tich m huang dan gidi

Phuong trinh dao dong tai mpt diem khi c6 giao thoa:

u = 2 A cos d, - d , cos (Ot - 71

1 ° J,"

Phuong trinh dao dong tai O:

u = 2Acos cot 27ta (voi / = 2a) Phuong trinh dao dong tai M:

C a u 7: Hai nguon song ket hop, dat tai A va B each nhau 20cm dao dgng theo

p h u o n g trinh u = acos(cjt) tren mat nuoc, coi bien do khong doi, buoc song

X = 3cm Gpi O la trung diem ciia AB M o t diem nam tren d u o n g trung true

AB, dao dpng cimg pha v o i cac nguon A va B, each A hoac B mpt do^n nho

nhat la

A.12cm B.lOem C.13.5cm D.lScm

<Phdn tkh vd hucmg dan gidi

Bieu thiic song tai A, B la: u = aeoscot

Xet diem M tren trung true ciia AB:

A M = B M = d (cm) > 10 cm

Bieu thiie song tai M : U ; ^ = 2acos c o t - 27id

Diem M dao dpng ciing pha voi nguon k h i

— = 2 k T i = > d = kX = 3 k > 1 0 ^ k > 3 , 3 3

k^in = 4 d^i„ =4X = 4.3 = 12cm

Chpn dap an A

C a u 8 : Trong thf nghi^m giao thoa tren mat chat long voi 2 nguon A, B phat

song ket hop nguoe pha nhau Khoang each giiia 2 nguon la AB = 16cm Hai

song truyen d i c6 buoc song la 4em Tren d u o n g thang xx' song song voi

AB, each A B mot doan 8cm, goi C la giao diem cua xx' v o i d u o n g trung true

eiia AB Khoang each ngan nhat t u C den diem dao dpng v o i bien dp cue

dai nam tren xx' la

A l,42em B l,50em C 2,15em D 2,25cm

^hdn tich v>d hucmg dan gidi

Cach 1 : Goi M la diem thoa man yeu cau va dat C M = x,

Khoang each ngan nhat t u C den diem dao dpng v a i bien dp eye tieu nSm

tren xx' thi M thupe cue tieu t h u nhat k = 0

X = 1,42 cm

!U,

Ban CO the giai theo phuong trinh hypeebol n h u sau:

a Trong do N la dinh hypeebol v o i

duong cue tieu gan trung true nha't ket hop voi hai nguon eung pha nen

voi e la tieu

= 1

O N = a = - = - = lem

4 4 diem va e = OB = O A = — = — = 8em

2 2 Taco: b^ =e^ - a ^ - 8 ^ - 1 ^ =63 Theo bai ra thi khoang each giua xx' voi AB la y = 8cm

dpng eung pha Bie't song do moi nguon phat ra eo tan so'f = lO(Hz), van toe truyen song 2(m/s) Gpi M la mot diem nam tren d u o n g vuong goc voi A B tai A dao dong voi bien dp cue dai Doan A M c6 gia tri Ion nhat la:

A 20em B 30cm C 40em

^han tich vd hu&ng ddn gidi

Taco ; = ^ = ^ = 20(em)

f 10 ^ '

Do M la mot eye dai giao thoa nen

de doan A M eo gia trj Ion nhat thi

M phai nam tren van eye dai bae 1

n h u hinh ve va thoa man:

T h a y (2) v a o (1) ta d u o c : SJAO^ + d ,^ - d ] = 2 0 : • d i = 3 0 ( c m )

C h p n d a p a n B

Cau 10: T r e n b e m a t chat l o n g c6 h a i n g u o n k e t hgrp A B each n h a u 100cm

dao d o n g c u n g p h a Biet s o n g d o m o i n g u o n p h a t ra c6 t a n so f = 10(H2),

v a n toe t r u y e n s o n g 2(m/s) G o i M la m o t d i e m nam t r e n d u o n g v u o n g goc

v o i A B tai A d a o d o n g v o i b i e n d p cue d a i D o a n A M c6 gia t r j n h o n h a t la:

Cau 11: T a i h a i d i e m A , B tren m a t chat l o n g c6 h a i n g u o n p h a t song:

Uys^ = 5cos((ot) c m ; U g = 3cos a)t + —

each h a i n g u o n n h u n g k h o a n g d i = 12,75>i v a d 2 = 7,25X se c6 b i e n d p d a o

d o n g la b a o n h i e u ?

A 7,5 m m B 1 m m C 15 m m D 56 m m

^hdn tick v>d hucmg ddn gidi

Do hai n g u o n eo b i e n d o khac n h a u nen ta t i n h bien d o s o n g t o n g h p p tai M

theo c o n g t h u e t o n g h o p dao d o n g d i e u hoa:

Chnyto 3

S < ^ N G D C T N G

A K I E N THLTC C a B A N

1 S\f p h a n xa cua song

K h i song gap mot vat can hoac diem cuoi cua m o i t r u o n g c6 song truyen

toi thi it nhat mot phan cua song bj phan xa lai

a) Phan xa ciia song tren vat can co d i n h

Thuc nghiem cho thay, k h i phan xa tren vat can co djnh bien dang cua

day b i dao chieu

Tqi sao khi phan xa tren vat can codinh, bien dang bj dao chieu?

Q

a)

b)

K h i bien dang (xung) toi dau Q, no tac d u n g mot luc hirong len vao gia

d o (tuong) Theo dinh luat 3 niuton, gia d a tac d u n g m g t luc bang va

nguoc chieu vao day Phan luc nay sinh ra mot xung tai gia da X u n g nay

truyen tren day theo chieu ngugc vai xung toi Trong phan xa nay, phai

CO mot nut tai gia do, v i day bj g i u co djnh tai do X u n g toi va xung phan

xa trai dau nhau, de chiing triet tieu Ian nhau tai d i e m do

Vay, khi phan xa tren vat can codinh, song phan xa luon ngUQC pha vai

song toi a diem phan xa

h) Phan xa cua song tren vat can t y do

Thuc nghiem cho thay, khi phan xa tren vat can t u do bien dang cua day

khong bj dao chieu

Tai sao khi phan xa tren vat can tu do, bii^'n dang khong bi dao chieu?

Vay, khi phan xa tren vat can tu do, song phan xa luon ciing pha vai song tai a diem phan xa , , ;

2 Song d u n g

a) D i n h nghia: Song truyen tren sai day dan hoi trong truang hap xuai hi^

cdc nut vd cdc bunggoi la song ditng

Song d u n g dugc h i n h thanh la ket qua ciia su giao thoa song toi va song phan xa N h i r n g diem tren day, tai do cac song triet tieu nhau thi khong

dao dong va dugc ggi la nut N h u n g diem tai do cac song dong pha vai nhau thi dao dong v o i bien dp cue dai va dugc ggi la bung

N h u n g nut va bung xen ke, each deu nhau

b) G i a i thich s u tao thanh song d i m g tren day

Xet dao dong ciia m o t phan •

Song to^i

;<_ d - -

tu tai diem M tren day each dau CO djnh B mot khoang

M B = d Gia s u vao thai diem

t, song toi deh B va truyen toi do mgt dao dong co phuong trinh dao dong la: Ug = A cos cot = Acos27ift Chgn goc tga do tai B, chieu duong la chieu tu B den M

->

M

Song phan xg

Phuong trinh song toi tai M : u^, = A cos 0)t + - 27td^

Phuong trinh song phan xa tai B: u g = - A cos (cot) = Acos(27tft - 7i) Phuong trinh song phan xa tai M : u ^ = A cos 27tft - 71 - 27td^

N h u vay, tai M dong thoi nhan dugc hai dao dgng cung phuong, cimg tan so Do do, dao dgng tai M la tong hgp ciia hai dao dgng do song toi

va song phan xa truyen den:

u = U ] ^ + U M = A C O S 27lft + — + Acos

u = 2 A c o s 27ld 7C + —

X 2

cos

Bien do song tai M : A^^ = 2 A cos 27:d

Bien do nay phu thuoc i<hoang each d t u M cten B (dau co'djnh ciia day)

+ Neu tchoang each d = thi bien do dao dong tai M bang 0, tai M c6

mot nut

+ Neu khoang each d = k + — — thi bien do dao dong tai do dat gia tri

V 2 / 2 eye dai, a dcS c6 mot bung song,

c) Khoang each giua nut va bung trong song dung

- Hal nut Hen tie'p each nhau mot khoang bang

- Hal bung hen tie'p each nhau mot khoang bang ^

- Khoang each giua bung scSng va nut song lien ke bang ^

3 Dac diem ciia song dung

a S u truyen nang lugng trong song dung

Tai sao goi la song dung? Co phai do nang lugng khong truyen di ma

dung lai? Neu d u n g lai thi d u n g lai o dau? Neu c6 truyen di thi truyen

nhu the nao? Ta hay xem xet van de nay

Nut luon ctiing yen nen no khong thuc hien cong Do do nang lugng

khong truyen qua dugc nut Burij^ khong bien dang, sue cang day tai

bung bang 0, ncn hun;^ cung khong thuc hien cong Do do, nang lugng

cung khong truyen dugc qua bung N h u vay, nang lugng ciia m o i doan

day dai hang 1/4 buoc song c6 mot dau la nut, dau kia la bung thi khong

doi N o i each khac, nang lugng " d u n g " trong moi doan day nhu vay

Nang lugng cua moi doan day la khong doi, khong c6 su truyen nang

lugng t u doan day nay sang doan day kia

Trong moi doan day c6 su truyen nang lugng khong? Co su bien doi t u

dong nang sang the nang khong? Co su truyen nang lugng t u diem nay

sang diem khac cua moi doan day ma ta xet Neu trong 1/4 chu k i , nang

lugng truyen tir trai sang phai vi du t u niit tai bung thi trong 1/4 chu ki

tie'p theo nang lugng truyen t u phai sang trai, t u bung toi nut

Tom lai: Nang lugng " d u n g " trong moi doan day dai 1/4 buoc song c6 mot

dau la nut, mot dau la bung Nang lugng khong truyen ra khoi doan day

cung nhu khong truyen vao doan day qua nut va bung Mat khac, trong moi

doan day thi nang lugng lai truyen qua lai t u dau nay toi dau kia, dong thai

CO su chuyen doi qua lai giua dong nang va the nang Vi the^ khi xet ve su bao toan nang lugng, doan day tuong duong con ISc 16 xo

Sv dao dgng cua cac diem tren day khi c6 song dirng

Tren doan day, trong dieu kien li tuong, cac nut hoan toan d u n g yen, cac diem con lai van dao dong voi van toe dao dong (can phan biet dugc toe

do dao dong cua phan tir moi truong voi toe dot truyen song)

T Thai gian giua hai Ian sgi day duoi thang lien tie'p la — (Khi day duoi thang, li do cua bung = 0 Thoi gian giira hai Ian lien tie'p li do bung song

= 0 la mot nira chu ki) i

c Tinh tuan hoan ciia song dirng ^

* Tinh tuan hoan theo khong gian: ^ , ,

-Bien do cua phan t u vat chat tai mot diem khi CO song dung: ; _

Af^^ = 2 A cos 27id n

+ —

X 2

Vay, ta coi bien do ciia phan t u moi truong dao dgng dieu hoa voi chu ki X

Trong truong hgp chi xet rieng bien do, no c6 the c6 gia tri am hoac duong, nhung khi xet chung voi phuong trinh song d u n g thi bien do luon duong

Khi chi quan tam toi bien do, ta d i m g thuat ngu : "Dp l^ch pha bien dp"

dao ctgng trong song dung N h u vay, tren sgi day dan hoi dang c6 song

dung on dinh each nhau mot khoang d thi dp l^ch pha bien dp la:

2nd

Acp - •

X

Cong thiic tren se rat tien Igi trong viec tinh bien do tai mot diem tren

day dang c6 song dirng khi biet khoang each tir no toi diem nut hoac

bung K h i xet tai dp lech pha bien dp nay ta khong can quan tam toi som

pha hay tre pha vi dieu ta quan tam la do Ion ciia bien do dao dgng

T i n h tuan hoan theo thai gian:

Phuong trinh song dirng tai mot diem:

2 A cos 2 7 l d , _^7t" > 0 t h i M , N cung pha dao

dong nghia la phuang trinh bien do mang cung 1 dau (tuc ciing am hoac

cimg duong) thi chiing dao dong cung pha

+ K h i 2 A cos 27tdi n - + —

X 2

2 A cos 27id < 0 thi M , N ngugc pha

dao dgng nghla la bieu thuc bien dp trai dau nhau

- V i tri cac diem dao dpng cung pha, ngupc pha

+ Cac diem doi xiing qua mpt bung thi dong pha (do'i xiing vol nhau qua

duong thSng di qua bung song va vuong goc voi phuang truyen song)

Cac diem doi xung vai nhau qua mpt niit thi dao dpng ngupc pha

M , P do'i xung qua bung B nen ciing pha

dao dpng De thay phuong trinh bien dp cua

M va P cimg dau Suy ra, M va P dao dpng

ciing pha

M , Q doi xung qua niit N nen ngupc pha

dao dpng De thay phuong trinh bien dp ciia

M va Q ngupc dau nhau Suy ra M va Q dao

dpng ngupc pha

Cac diem thupc cimg mpt bo song (khoang giiia hai nut lien tiep) thi dao

dpng cimg pha v i tai do phuong trinh bien dp khong doi dau Cac diem

nam 6 hai phia ciia mpt nut thi dao dpng ngupc pha v i tai do phuang

trinh bien dp doi dau khi qua nut

Mpi diem thupc bo song NimN2 c6 phuang trinh bien dp mang cimg dai'

Cty TNHH MTV DWH Khang Vift

duong

Vay, chiing dao dpng ciing pha

Cac diem thupc hai bo song lien tiep

NimN2 CO bien dp mang dau duong va

bo song N2nNi c6 bien dp mang dau am

Tir do M va P dao dpng ngupc pha

4 Dieu ki^n de c6 song dimg tren spi day dai /:

* Hai dau la nut song: 1 = ( k e N * )

So bung song = so bo song = k

Vi d u 1: (Trich de thi thu chuyen Ha Tinh Ian 1 nam 2013)Tren mpt spi

day CO song dirng voi buoc song la A, tren day quan sat thay 4 bung

song Khoang each giua niit song thu 2 den bung song t h u 4 la

A 1/2 B.2.A C.5A/4 D A

^hdn tick vd huong dan gidi

Theo bai ra ta c6 hinh ve:

Ta xem A la nut thir nhat vi the khoang each giira niit song t h u 2 deh bung

song t h u 4 la: A + — = —

4 4 Chpn dap an C

Bi quyei on luy^n thi dai hgc dat diem lot da Vat It, tap 1-Le Van Vinh

Vl dM 2: (Trich de thi thu chuyen Ha Tinh Ian 2 nam 2013) Song truyen

tren sgi day dai / vai budc song X, de c6 song dung tren sgi day vai mpt

dau day c6' djnh va mpt dau day tu do thi ><

<PMn tick v>d hu&ng ddn gidi

Chieu dai spi day mpt dau co djnh,

mpt dau tu do de c6 song dung la:

Chpn dap an A

Nh^n xet: denha cong thiec tren doi khi de dang

nhimg rat de nham Ian Khi su dung hinh ve

chung ta se c6 dugc cong thitc tren nhanh va

chinh xdc horn nhieii, sau nay khi giai cdc bai

todn ve tinh nut, bung, bo song hay chieu dai

tuang ung tht sie dung hinh ve Id tot nhat

Tu hinh ve ta c6 chieu dai day thoa man:

Vl dv 3: (Trich de thi thu chuyen d?i hgc vinh Ian 3 nam 2012)

Mpt spi day dan hoi AB hai dau c6' dinh dupe kich thich dao dpng voi

tan so 20Hz thi tren day c6 song dung on dinh vai 3 niit song (khong

tinh hai nut 6 A va B) De tren day c6 song dung voi 2 byng song thi tan

so' dao dpng ciia spi day la

A 10 Hz B 12Hz C.40HZ D 50 Hz

^hdn tick vd huang ddn gidi

Theo bai ra: khi tan so f = 20Hz thi song dimg c6 5 nut ke ca A va B nen c6 4

bo song Vay chieu dai day thoa man: \ k^ = 4.^-2X = ^ (1)

Khi tan s o / thi song dung c6 2 byng song nen c6 2 bo song Vay chieu dai

day luc nay: 1 = k ' y = 2.j ^X' = ^ (2)

T u ( l ) v a (2) taco: ^ = ^ => f = | = ^ = lOHz Sau day la hinh ve bieu dien qua trinh tren:

Cty TNHH MTV DWH Khang Viet

Chpn dap an A

Vl dM 4: (Trich de thi thu chuyen dai hpc Vinh Ian 3 nam 2012)

Mpt spi day dan hoi dai 60cm, toe dp truyen song tren day 8m/s, treo lo limg tren mpt can rung Can dao dpng theo phuong ngang voi tan so f thay doi

tu 80Hz deh 120Hz Trong qua trinh thay doi tan s6^ c6 bao nhieu gia tri tan

so CO the tao song dung tren day?

A 15 B 8 C.7 D.6

i'hdn tick m huang ddn gidi

Chieu dai day thoa man:

Theo de bai: 80Hz < f < 120Hz o 80 < — (2k +1) < 120 11,5 < k < 17,5 => k = 17 -12 +1 = 6

_ V a y CO 6 gia tri cua tan so cho song dung tren day Chpn dap an D

Vl du 5: Trong thi nghi^m ve song dimg, tren mpt spi day dan hoi dai 1,5 m voi hai dau co dinh, nguoi ta quan sat thay ngoai hai dau day

CO djnh con co nam diem khac tren day khong dao dpng Biet khoang

thoi gian giua ba Ian lien tiep voi spi day duoi thMng la 0,1s V?n toe

truyen song tren day la

_ A 5m/s B 4m/s C lOm/s D 6 m/s

Biquyet on luyjn thi dai hoc dat diem td'i da Vat It, tap 1 - Lc Van Vitih

'Phdn tich vd hiiong ddn gidi

Khoang thai gian giua ba Ian lien tiep soi day duoi thang la mot chu ky:

At = T = 0,]s=>f = 10Hz

Ngoai hai dau day co djnh con c6 nam diem khac tren day khong dao

dong CO tat ca 7 nut song => c6 6 bung song => c6 6 bo song =i> k = 6

Chieu dai soi day thoa man: 1 = k - = 6 - = = 3 ^ => v = = - ^ ^ ^ = 5m/s

^ 2 2 f 3 3

Chpn dap an A

Vl du 6: (Trich de t h i t h u chuyen Ha TTnh Ian 2 nam 2013)

Khi CO song d u n g tren spi day dan hoi AB voi dau A la diem nut va dau

B la diem bung thi

A diem tren day each dau A mot doan b^ng nua buoc song la diem bung

B diem tren day each dau A mot doan bSng mot phan t u buoc song la diem nut

C diem tren day each dau B mot doan bang ba phan tu buoc song la diem bung

D diem tren day each dau B mpt doan bang mot phan t u buoc song la

diem nut

^kdn tich vd himng dan gidi

A (sai) vi diem tren day each dau A

mot doan bSng nua buoc song

khong phai la bung song ma la nut

song

B (sai) vi diem tren day each dau A

mpt doan bSng mot phan t u buoc

song khong phai la nut song ma la

bung song

C (sai) VI diem tren day each dau B

mpt doan b3ng ba phan t u buoc

song khong phai la bung song ma

la niit song

D (dung) V ! diem tren day each dau B

mot doan bSng mot phan t u buoc

song la diem nut

H i n h ve tren se cho cai nhin cu the hon

Niit

p B A I T A P VAN DgNG:

C a u 1: Mot sgi day AB = 50cm tree lo lung dau A co djnh, dau B dao dpng voi tan so 50Hz thi tren day co 12 bo scSng nguyen K h i do diem N each A mot doan 20em la bung hay nut song t h u may ke t u A

A bung song t h u 6 B nut song t h u 6 f;

C byng song t h u 5 D nut song t h u 5 r

^hdn tich vd hu&ng ddn gidi ^," - '

Tren day CO 12 bo song nguyen nen k = 12 ,; ,

Chieu dai sgi day mot dau t u do, mot dau co djnh khi co song dung thoa man:

A Co, c616 bung song B Co, co 21 bung song

C Khong D Co, CO 15 bung song

^hdn tich vd huang ddn gidi

Bai toan chua cho bie't song dirng tao ra trong truong hop nao nen ta xet ca hai truong hpp: i »•

Truong hpp 1: Hai dau day e o d m h Chieu dai day thoa man: l = k - = k — ^ k = — = ^-^'^-^^ = 1 6 e Z

Suy ra co song d u n g xay ra ung voi 16 bung song

Do truong hpp 1 thoa man nen ta khong xet truong hpp 2 nira Truong hpp

2 la song d u n g tao ra tren day mot dau co dinh, mpt dau t u do

Chpn dap an A

C a u 3: M o t spi day dan hoi dai 130cm, co dau A co' djnh, dau B t u do dao dong voi tan 50 Hz, van toe truyen song tren day la 20 m/s Tren day co bao nhieu niit va bung song:

A CO 6 nut song va 6 bung song B co 7 niit song va 6 bung song

C CO 7 nut song va 7 bung song D co 6 nut song va 7 bung song

^hdn tich vd huang ddn gidi

Nhan xet: song dirng tao ra boi mpt dau t u do, mpt dau co dinh

243

=> so niit = so bung = so bo + 1 Vi the loai dap so B va D (do bung, nut

khong bang nhau)

Chieu dai soi day thoa man:

Chpn dap an C

Cau 4 : Soi day AB = 35cm voi dau B tu do Tao ra tai A mot dao dong ngang

CO tan so f Van toe truyen song la 3m/s, muon c6 10 bung song thi tan so

dao dgng phai la bao nhieu ?

D^ng 2: Xac djnh dieu k i f n de c6 song duTng t r e n day

Vl d u 1 : Soi day dan hoi AB, dau A gan voi can rung c6 tan so f, dau B

dugc giu CO dinh f i va ii la hai tan so lien tiep de tao ra song dung tren

soi day Tim tan so nho nhat de tao ra duoc song dimg tren sgi day

^hdn tick m hu&ng dan gidi

Dieu kifn de c6 song dung: AB = / = k.-^ = k.-;^ ^ ^ = (1)

Z Z t Z/ K

Hai tan so fi, h la hai tan so lien tiep de tao ra song dirng ung voi so bo

song Ian lugt la k va k +1

Vl dM 2: Sgi day dan hoi AB, dau A gan vai can rung c6 tan so f, dau B tu

do fi va h la hai tan so lien tiep de tao ra song dirng tren sgi day Tim tan

so nho nha't de tao ra dugc song dung tren sgi day ,j j • ,

'Phdn tick va hu&ng dan gidi

Dieu kien de c6 song dung:

AB = / = (2k + l ) - = (2k + l ) — — = — ^ (1)

^ ^ 4 ^ ^ 4f 41 2k + l ^ ^

Hai tan so fi, k la hai tan so lien tiep de tao ra song dimg ung voi so bo

song Ian lugt la k va k + 1

Chieu dai sgi day mot dau tu do, mot dau codinh khi c6 song dimg thoa man: l=k—+—=

2 4 k + - 2, k + - 2 Dat: fk = k + -

C _ V _ fk+uJi — Suy ra dap an van la C

V l d u 3 : M o t soi day dan hoi duoc treo thSng dung vao mot diem co dinh,

dau con lai tha t u do N g u o i ta tao ra song d u n g tren day v o i tan so nho

nhat la ^ Phai tang tan so them mot lugng nho nhat la bao nhieu de lai

CO song d u n g tren day? j

A Af = ^ B Af = 2 ^

^hdn tick pd hiccmg dan gidi

Chieu dai soi day mot dau t u do, mot dau co dinh khi co song dung thoa man:

(2)

( l ) v a (2)=> f = ^ = 3=>f, = 3 ^ = ^ A f = f i - ^ - 2 ^

41 Vay phai tang them Af = 2^, thi tren day lai c6 song d i m g

Chpn dap an B ^

V i du 4 : N g u o i ta tao song dung trong mot cai ong mot dau kin mot dau

ho dai 0,85m chua day khong khi 6 dieu k i ^ n thuong (van toe am la

340m/s) H o i tan so nho nhat de co song dung trong ong la bao nhieu?

A f - 50 H z B f = 75Hz C f - 2 0 0 H z D f = 100Hz _

^hdn tick m hu&ng d&n gidi

Truong hop nay giong truong hop song dung tao ra boi sgi day mot dau co

dinh, mot dau t u do

Dau CO d j n h (dau k i n cua ong) la nut

Dau t u do (dau ho cua ong) la bung song

=> Chieu dai ong day thoa man:

, T»}/ nhien, ncu dcy chung ta sc thai/ dugc trifdng hap giao thoa cua song dimg tao

ra trong ong mot dau kin, mot dau ha giong truang hap song dimg tren sai day

mot dau CO djnh, mot dau tie do VI thechi can dp dung cong thitc tinh tan sonho

nhat cua trmrng hap nay Id co ngay dap dn nhanh nhat

340

f = — = 'min

vi de cho van toe truyen song va chieu dai day nen ta co

= 100Hz :

41 4.0.H5

m B A I T A P VAN DUNG:

Cau 1: M p t soi day dai 1 = 1,2 m co song dung voi 2 tan so lien tiep la 40 Hz va

60 Hz Xac djnh toe do truyen song tren day biet hai dau day deu la nut song?

A 48m/s B 24m/s C 32 m/s D 60 m/s

^hdn tick m hu&ng dan gidi

Dieu kien de co song dung tren day v o i hai dau day la m i t song (hai dau co

T~ = r^- K h i f, va (2 la hai tan so lien tiep fj < £3 thi k j va k j la 2 so

nguyen lien tiep: k2 = k| +1

Tuy nhien, ne'u ndm dugc cong thitc a vi du 1 thi ket qua thu diegc se nhanh han

Tan so nho nha't de co song dung tren day (hai dau co djnh):

^min=Y^=h-h-^^ = 2/(f2 - f i ) = 2.1,2(60 - 40) = 4 8 m / s

247

C a u 2: M o t s(?i day dan hoi c6 mot dau co djnh, mpt dau t u do, tren day c6

song d u n g v 6 i hai tan so Hen tiep la 30Hz, 70Hz T a n so n h o nhat de c6

song d u n g tren day la

A f„.„ = 30Hz B f^,„ = lOOHz C f„i„ = 20Hz D f^,„ = 70Hz

^hdn tich \pd hitomg ddngidi ,\

Tan so nho nhat de c6 song d u n g tren day (mot dau t u do, mpt dau co dinh)

f n n = ^ ^ = ^ = 2 0 H z C h p n C

C a u 3 : M p t spi day dan hoi dupe treo thSng d u n g vac mpt diem N g u o i ta

cho diem nay dao dpng, tao ra song d u n g tren day v o i tan so nho nhat la fi

De lai co song d u n g , phai tang tan so'toi thieu den fi T i so j - bang:

A 1/4

'2

D 3

B 1/3 C 4

'Phdn tich ra huang dan gidi

Spi day dan hoi dupe treo t h i n g d u n g vao m p t diem ta hieu rang diem con

lai dupe tha t u do v l the'trong t r u o n g hop nay chieu dai spi day thoa man:

C a u 4 : Cho ong sao co mpt dau bjt k i n va mpt dau de ho Biet rang o n g sac

phat ra a m to nhat u n g v a i hai gia trj tan so cua h a i hpa a m lien tiep la

lOOHz va 280Hz Tan so am nho nhat k h i ong sao phat ra a m to nha't bang

A 50 H z B 4 8 H Z C 75 H z D 90 H z

^hdn tich vd huang dan gidi

Day la t r u o n g h o p tao song d u n g cua song am T r u o n g h p p nay giong

t r u o n g h p p song d u n g ciia spi day mpt dau t u do, m p t dau co'djnh

, r u - c - f k 280-100

C h p n dap an D

p a n g 3 : U'ng dung khai n i e m If ch pha bien do d e giai toan

A K I E N THU'C CAN NAM:

1 Bien d p ciia phan t u m o i t r u o n g co p h u o n g trinh:

A M = 2 A C O S 27id

2 H a i d i e m tren day dan hoi dang co song d u n g o n d j n h each nhau khoang

d thi d p I#ch pha b i e n dp dao dpng la: Aep = — ^

K h i su d u n g khai niem ve d p lech pha bien d p de t i n h bien d p k h o n g can quan tam toi khai n i ^ m som pha hay tre pha

B B A I T A P VAN DUNG:

Vl du 1: Tren m p t spi day dan hoi dang co song d u n g o n d m h , B la mpt

b u n g song, bien dp dao dpng tai b u n g la A D i e m M each B m p t doan

d i i n g bang — T i n h bien d p dao dpng tai M

A A M = 2 A C. A M = A D - A M = A V 3

<?/idn tich ra huang ddn gidi

Dp l^ch pha bien dp dao dpng giira M va B:

Bien d p dao dpng trong song d u n g co the

am hoac d u a n g hoac bang k h o n g tuy nhien trong giai han ciia chucmg trlnh, ta chi xem

A

bien d p la d u o n g v i the A ^ = —

C h p n dap an B

Vl d u 2: Tren mpt spi day dan hoi dang co song d u n g o n dinh, N la mpt

nut song, bien d p dao dpng tai b u n g la A D i e m M each N m p t doan

d i i n g hang — T i n h bien dp dao dpng tai M

*5

A A M = 2 A B. A M = ^ C A M = A D. A M = A V 3

249

<Phdn tich vd huang dan gidi

D o lech pha bien dp dao dpng:

X

Acp = 2nd _ ^ " 3 _ 2n

X X 3

Sit dung vec to quay nhu hinh ve:

De dang thay rang, bien do dao dong

tai M bang —^—

Chpn dap an D

V l du 3: (De thi thu T H P T Chuyen Phan Bgi Chau - Ngh? A n Ian 3

nam 2012)

M , N , P la 3 diem lien tiep nhau tren mot soi day mang song d u n g c6

cung bien do 4mm, dao dong tai N nguoc pha voi dao dong tai M

N P = 2 M N = 2cm C u sau khoang thoi gian ngSn nhat la 0,04s soi day c6

d^ng mpt doan thMng Toe dp dao dpng cua phan ti> vat chat tai diem

bung khi qua vj tri can bang (lay n = 3,14)

A 375mm/s B 363 mm/s C 314mm/s D 628mm/s

<Phdn tich m huang dan gidi

M , N dao dpng ngupc pha, cung bien dp dap dpng nen chiing dpi xiing

nhau qua mot niit song

N , P cung bien dp vi the N va P doi xung nhau

qua bung nen N , P dao dpng cung pha Vay M va

P dao dpng ngupc pha nhau =t> M P = n

Vay bung spng CP bien dp: A g = = 2 A N = 2.4 = 8 m m

CPS

-3

T Thoi gian giiia hai Ian day dupi thSng lien tiep la — = 0,04 T = 0,08s

Cty TNHH MTV DWIi Khang Vi?t

Toe dp cue dai cua diem bung: v

Chpn dap an D

max = M A B = ^ A B = - ^ 8 = 6 2 8 m m / s

0,08

Vl du 4: Tren soi day thang CP spng dung, khoang each giira mot nut va nut

thu 4 ben phai no la 15 cm Dp lech pha giua hai diem M , N ( M khong triing vai nut song) tren day each nhau 1,875cm c6 the c6 gia trj bSng gia tri nao trong cac gia tri sau :

A 7T/8 rad B 3TI/4 rad C n/2 rad D n rad

'Phan tich hirnng dan gidi

Nhan xet: )icii vao each ^idi thi ta klwn^^ thay c6 van dc sal cd, dap an thu

dugc CO twn^ ban dap an dc cho Lieu ci'nis^ thirc tinh do tech pha ^ifm hai diem tnvi^; qua tnnh trui/en son;^ cd khdc cdn^ thur tinh do lech pha twng song dirng

hay khong? Co mot dieu cac ban nen nha la: tai hai diem tren sai day cd aong

dimg thi chi cd the la ciing pha hoac ngiegc pha Cac diem cd bien do ciing dd'u thi Cling pha vd iiginrc Iqi, cac diem cd bien do khdc dd'u thi nguvc pha Vi theta khong tlie sir dung cdiig thur tinh do lech pha ciia hai diem trong moi tritvng cd song

truyen qua dp dung vdo sdng dimg duw Cdcli gidi tren Id sai hdn vebdn chat vi

the cac ban can lint 1/ khi Idm triic nghiem

_yay M va N chi co the cung pha hoac ngupc pha Chpn dap an D

V i du 5: (Dai hpc2011):

Mot soi day dan hoi cang ngang, dang co song d u n g on djnh Tren day, A

la mot diem nut, B la mpt diem bung gan A nhat, C la trung diem cua AB, vpi AB = 10 cm Biet khoang thoi gian ngan nhat giOa hai Ian ma li do dao dpng cua phan t u tai B bang bien dp dao dpng cua phan t u tai C la 0,2 s Toe dp truyen song tren day la

A l),25m/s B 2m/s C 0,5 m/s D 1 m/s ' '

(Phdn tich vd hitcmg ddn gidi

De dang tinh dupe buoc song: A = 4.AB = 40cm

Dp lech pha bien dp dao dpng giOa B va C:

\(PBC = 271

Bung B

N i i t A

Bi quye't on luy^n thi dai hgc dat diem tot da Vat U, tap 1 - Le Van Vinh

Ta c u n g c6 the t i n h c h u k y theo v o n g t r o n l u p n g giac n h u sau:

T h a i g i a n ngan nha't de l i d p dao d p n g cua p h a n t u t a i B b a n g b i e n d p dao

d p n g tai C u n g v o i goc quet 2.45° = 90"

Vay © = - = ^ = 2 , 5 7 i ( r a d / s ) = — = l , 2 5 ( H z )

t 0 , 2 ^ ' 2n

v = ? f = 40.1,25 = 5 0 ( c m / s )

T a cung c6 the giai bai toan tren theo p h u o n g trinh song n h u sau

Ta C O b i e n d p song d u n g tai m p t d i e m M tren day, each d a u c o ' d i n h A doan

N h a n x e t : khi sit dung do lech pha bien do de gidi bai toan tren thi bai toan tren

khd giohg vai bai toan ve thai gian trong dao dong dieu hoa vi these de hieu va

ngan gon han nhieu so vai each giai khdc

V I d u 6 : M p t sai day dan h o i cang ngang, dang c6 song d u n g o n d i n h Tren

day, A la m p t d i e m niit, B la d i e m b y n g gan A nha't v o i A B = 18cm, M la m p t

d i e m tren day each B m p t khoang 12cm Biet rang trong m p t chu k y song/

khoang t h a i gian m a d p I o n van toe dao d p n g cua phan t u B n h o h a n v a n toe

eye dai ciia phan t u M la 0,1s Toe d p t r u y e n song tren day la:

A 3,2m/s B 5,6m/s C 4,8 m/s D 2,4 m/s

Cty TNHH MTV DWH Khang Vi^t

'Phan tich vd hu&ng dan giai

Toe d p eye d a i cua p h a n t u m o i t r u o n g

tai b u n g song la: Vg^ax = ' ^ • 2 A

d a n g C O song d u n g Q u a n sat tren d a y tha'y c6 cac d i e m each d e u n h a u

n h i i n g k h o a n g 10cm l u o n dao d p n g c u n g bien d p n h a u So n u t song tren day la:

^hdn tich v>d huang dan gidi

Gpi M , N , P, Q, R la cac d i e m lien tiep n h a u dao d p n g c u n g b i e n d p Theo de eho ta c6: M N = N P = PQ = QR =10 c m

Theo h i n h ve b u o c song: = M R = 4 M N = 4.10 = 40em = 0,4m | Chieu d a i spi d a y h a i d a u co d i n h thoa m a n : j t ,.i, ;|

Bi qiiyet on luyfit thi itai IIQC dat diem td'i da Vat It, tap 1 - Le Van Vinh

C a u 1: Tren day AB co song dung voi dau B la mot nut Song tren day c6 bu6c

song A Hai diem gan B nhat co bien do dao dong bSng mot nua bien dg dao

dong cue dai cua song dung each nhau mot khoang la:

A A/3; B.A/4 C A/6; D A/12;

'Phdn tich ra hii&ng dan gidi

Ggi C la bung gan nut B nhat n h u hinh ve va M , N la hai diem co bien do

dao dong b5ng mot nua bien do dao dgng cue dai (bien dg dao dong cua

diem C)

A ,

T u hinh ve ta co: cosCOM = r = » => C O M =

-Tuong t u ta cijng co:

Ac-ihco tilth tiiifii hoan cua song theo khong gian va thai gian ta co:^ <^ ^

Viiy M N = C(7c7! mn/ nhanh va tirang doi dchicu nen cac ban chi'i y

C a u 2: Tren mot sgi day cang ngang voi hai dau co djnh dang co song dung

Khong xet cac diem bung hoac nut, quan sat thay nhi>ng diem co cung bien

do va 6 gan nhau nha't thi deu each deu nhau 15cm Tan so dao dgng tren

day la 50 H/ Van toe truyen song tren day co gia trj b5ng

A 30em/s B 30 m/s C 90 cm/s D 45 m/s

<Phdn tich vd hitang dan gidi

Ggi M , N , P, Q, R la cac diem lien tiep nhau co ciing bien dg va each deu nhau

Theo de cho, ta co: M N = N P = PQ = QR =15 cm

:=> Buoc song: A, = M R = 4 M N = 4.15 = 60em ' Van toe truyen song tren day: v = A,f = 60.50 = 3000cm / s = 30m / s

Chpn dap an B Cau 3: M o t song d u n g tren mgt doan day co budc song bang 30em va bien

dg dao dgng eiia mgt phan t u each mgt niit song mgt doan 5cm co gia tri la

9 m m Bien dg A cua bung song la:

i'hdn tich i?d hitong dan gidi

Ggi N la niit va B la bung gan N nhat, diem each nut 5cm la M

Dg l^eh pha bien dg tai N va M :

C a u 4: Tren mgt sgi day dan hoi dang xay ra song dung, hai diem rieng bi^t

tren day tai mgt thoi diem khong the

A dao dgng ngugc pha

C dao dgng lech pha ^ •

^hdn tich vd hu&ng dan gidi

Xet A la niit va B la mgt b y n g song gan nhau nhat n h u hinh ve

Ggi M , N doi xung v o i nhau qua bung B suy ra

M , N dao dgng cung pha

Ggi N , Q doi xung voi nhau qua niit A suy ra N ,

Q dao dgng ngugc pha

Hai diem doi xwtg myi nhau qua mot bung thi dao dong cung pha

B dao dgng l^ch pha —,

3

D dao dgng ciing pha

Hai diem dot ximg vai nhau qua mot nut thi dao dgng nguac pha

N e u xet ve pha t h i tren m o t soi day dan hoi dang xay ra song d i r n g chi co

the c i i n g pha hoac n g u g c pha, k h o n g c6 v u o n g pha va cac pha khac

C h p n dap an C *

Cau 5: M o t soi day dan hoi cang ngang, dang c6 song d u n g on d i n h Tren

day N la m o t d i e m nut, B la m o t d i e m b u n g gan N nhat, N B = 25 cm, goi C

<Phdn tick vd hu&ng ddn gidi

N , B la n u t v a b u n g gan nhau nhat => N B = —

4

Buoc song: X = 4 N B = 4.25 = 100 cm

Theo bai ra: bien d o A ^ = ABV3

Do lech pha bien do tai B va C la:

Acp = 271 = — => BC = — = = — c m

X 3 6 6 3

C h p n dap an C

Cau 6: Mot soi day dan hoi cang ngang, dang c6 song dung on djnh Tren

day A la mot niit, B la diem bung gan A nhat, AB = 14 cm Bien do tai bung

^hdn tick vd hu&ng ddn gidi

Theo bai ra: Tren day A la m o t nut, B la d i e m

b u n g gan A nhat nen X = 4.AB = 4.14 = 56 c m

D o lech pha bien d p ciia A va C la:

14 A<p = 2 : r ^ = 2 K - 3 - = ^

1 Song am - cam giac am:

« Song dm la n h u n g dao d o n g co truyen trong cac m o i t r u o n g k h i , l o n g , ran

T r o n g chat l o n g va chat k h i song am la song doc, t r o n g chat r a n song a m

g o m ca song n g a n g va song doc , , ,

4 Phdn loai:

- Song am co tan so t u 16Hz den 20000Hz la n h i i n g dm nghe dwoc va

t h u o n g g o i la dm thanh

- Song a m co tan so n h o h o n 16Hz goi la ha dm i J i •

- Song a m co tan s o l o n h o n 20000Hz gpi la sieu dm

Tai n g u o i k h o n g nghe d u p e ha am va sieu am

N h i r n g a m co tan so xac d j n h {iiaxj do thi dao dong cua dm bien thicn tuan

hodn) t h i g p i la nhac am, con n h i r n g am co tan so k h o n g xac d i n h (hay do thi dao dong cua dm bien thien khong tudn hoan) t h i gpi la tap am

2 C a c dac tinh vat li: '

a Tan so am: T a n so a m bang tan so dao d p n g ciia n g u o n '

f b Cubng do dm - Muc cuang do dm:

v i I B = lOdB

L(dB) = 10.1g-^

^0

M u c c u a n g d p a m t h u o n g gap co trj

so vao k h o a n g t u 20dB den lOOdB

c Do thi dao dong am: T o n g h p p quy luat dao d p n g a m ciia tat ca cac hpa

am t r o n g m o t nhac a m cho ta do t h i dao d p n g a m cua nhac a m do D u n g

cu khao sat d o thj dao d p n g a m la dao d p n g k i d i e n t u , n g u y e n l y hoat

d p n g la bien dao d p n g a m t h a n h dao d p n g d i § n

3 Cac dac tinh sinh li:

a Do cao cua dm: Do cao ciia am la dac tinh sinh ly cua am do tan so' am

quye't dinh, am cao thi c6 tan so Ion hon am tram (thap)

b Am sac: A m sac do quy luat bien thien tuan hoan ciia cac hpa am (do thi

dao dong am) quye't dinh

c DQ to cua dm:

- 6 mot tan so xac dinh, cuong dp am cang Ion, cho ta cam giac nghe thay

am cang to Tuy nhien do to cm dm khong ti le v&i thuan voi cuang do am, ma

no con phu thugc vao tan so ciia dm

- De am thanh gay ra dupe cam giac am, cuong dp am phai Ian hon mot

gia tri cue tieu nao do gpi la nguang nghe Ngudmg nghe thay doi theo tan so

cua dm, v6if= lOOOHz thi ngudmg nghe la Um = L = 10 ' " W/m^ {con goi la

cuong do dm chuan)

- Khi cuong dp am len den lOW/m^ (ung voi muc cuong dp am 130dB) doi

voi mpi tan so cua am deu gay cho tai nguoi nghe cam giac nhuc nhoi,

rat kho chiu Gia trj nay ciia cuong dp am gpi la nguong dau

- Dp to ciia am phu thudc vao tan so dm vd cuang do dm hay muc cuong do dm

Dp to ciia am ung voi mot tan so xac dinh boi: AI = I - Imin. Mien nghe duac

ciia am nam trong khoang tu Imm den Imax

- Tai nguoi nghe dupe am c6 muc cuong dp am tu 0 (dB) den 130 (dB)

4 Cong suat ciia nguon am (nguon diem): P = S.I = 47cR^.I

(R la ban kinh mat song cau)

5 Neu nguon phat am la nguon diem phat am c6 dinh hudng theo Yanh

quat cau thi S la di^n tich chom cau: S = 27i.R.h

Vl du 1: (Trich de thi thu chuyen Ha Lpng - Quang Ninh Ian 1 nam 2013)

Ket luan nao khong dung voi song am?

A Toe dp truyen am trong moi truong ti 1§ voi tan so'am

B A m nghe dupe eo eung ban chat voi sieu am va ha am

C A m sac, dp cao, dp to, tan so la nhiing dac trung sinh ly eiia am

D Song am la cac song eo truyen trong cac moi truong ran, long, khi

Cty TNHH MTV DWH KffangViit 'Phdn tich vd hu&ng dan gidi

A (diing) vi toe dp truyen am trong moi truong la v = Xi ti le voi tan so am

va buoe song nhung chii y rang to'c dp truyen am trong moi truong khong phu thupc vao tan so am va buoc song ma chi phu thupe vao moi truong truyen am

B (diing) vi am nghe dupe c6 eiing ban chat vai sieu am va ha am chi khac nhau ve pho tan so' Cu the la: am nghe dupe c6 tan so' tu

16 < f < 20000(Hz), ha am eo tan so tu f < 16(Hz) va sieu am c6 tan so tu

f > 20000(Hz)

C (sai) vi am sac, dp cao, dp to la nhiing dac trung sinh ly ciia am eon tan

so'la dae trung vat ly ciia am

D (diing) vi song am la cac song eo truyen trong cac moi truong ran, long,

khi nhung khong truyen dupe trong ehan khong

Chpn dap an C

Vi du 2: Song am truyen trong khong khi den tai nguoi Tai c6 the cam nhan dupe am c6 ehu ky bang? , >:

A 0,2 s B 0,5^s C.lOms D 5 ns

'Phan tich vd hu&ng dan gidi

Tai ta cam nhan dime dm c6 tan so tit 16Hz den 20000Hz vi the chi can tim tan so ung voi timg chu ky decho vd sau do so sdnh vai tan soma tai ta nghe duac Id cd ngay dap an

! Voi T = 10ms = 10"^s => f = = = lOOHz => day la song am ma tai nguoi nghe dupe ^

D Voi T = 0,5ns = 0,5.10'^s => f = - = ^—Q- = 2.10^HZ => day la song

T 0,5.10'^ ^

sieu am 'iu-v

^ C h p n dap an C _ _ _ _

1 Vl du 3: Mpt nguon song am dupe dat trong nuoc Bie't khoang each giiia

mm hai diem gan nhau nha't dao dpng ciing pha nhau la 2,5m va van toe truyen

am trong nude la 1,8.10'^m/s Tan so'ciia song am do la

A 0,6kHz B 1,8kHz C 0,9kHz D 3,2kHz

Bl lJUyet on lUy^ tWl UUl ni^U uui uwm ivi uu vui n, lUff I—

'Phdn tich v>d huang dan gidi

Do lech pha cua hai diem dao dong cung pha: A(p = = kin

=> Ad = v a i k e N * => k ^ j ^ = 1 => Ad^^ir, = >^ = 2=>;^ = 2 m

V 1,8.10^ ' ? T t = ' : i l ' S "

Tan so cua song am: => f = — = - ^ - ^ — = 900Hz

Chpn dap an C

V l d u 4 : M o t nguon am dat tai O phat song am c6 cong suat khong doi

trong mot moi truong truyen am dang huong va khong hap thu am Hai

diem A, B each nguon a m Ian luQ-t la R i va R2 Biet cuong do am tai A gap

16 Ian cuong do am tai B Ti so — nhan gia trj nao sau day

A 4 B 16 C 1/16 D 1/4

^hdn tich v>d hu&ng ddn gidi

V l moi t r u o n g phat song am la dang huong nen song phat ra d u a l dang

hinh cau nen d i ^ n tich mat cau dugc xac dinh theo cong thuc: S = 47iR^

Ta biet rSng: cuong do am t i 1# nghjch voi binh p h u o n g khoang each t u

p p

diem can tinh cuong do am voi nguon am, cu the la: I = — = — r y

Theo bai ra, ta c6:

I A = 1 6 I B O — = 1 6 — SB = 1 6 S A O 47iR^ = 16.47tR?

R? R i

Chqn dap an A

Vl d u 5: M o t song am truyen trong khong k h i M i i c cuong do a m tai diem

M va tai diem N Ian luot la 50 dB va 100 dB Cuong d o am tai N Ion hon

cuong do a m tai M

A 100000 Ian B.1000 Ian C 50 Ian D.5000 Ian

'Phdn tich vd huong ddn gidi

M u c cuong dQ am tai M va N Ian lugt la: = 1 0 1 o g i ^ va L M = 1 0 1 o g y i

H i f u muc cuong dp am tai N va M :

L N - L M = 1 0 1 o g i ^ = 1 0 0 - 5 0 = 50 ^ ^ = 10^ ^ 1 ^ = I O ^ I M •

Chpn dap an A

cry iNHii Miv nvvn~Knang Viet

Vl d u 6: Nguon am O phat ra mot am c6 cong suat P khong doi, truyen dang huong ve moi phuong Tai diem M each O m o t doan 100m, m u c cuong do am la 100 dB Gia su m o i truong khong hap t h u am M u c cuong dp tai diem N each nguon mot doan 15m la:

A 130dB B 150,5dB C 170dB D 116,5dB

'Phdn tich vd huong ddn gidi

Theo de ra, ta c6: O M = 100m ,

Muc cuong dp am tai M va N Ian lupt la: = 1 0 l o g ^ va Lj^ = l 0 1 o g ^

Hieu muc cuong dp am tai A va B la:

[oMj = 10 log llOOy = -16,5

Vl d u 7: Tai mot diem A nam each xa nguon am N (coi n h u nguon diem) 1

khoang N A = 8 m eo muc cuong dp am la L A = 110 dB Biet nguong nghe

eiia am do la L = 10" W / m l Xet diem B nam tren d u o n g N A va each N khoang NB = 10 m Cuong dp am tai B la:

A 6.10^^ W / m l B 6,46.10"2 W/m2

C 10-2W/m2 D.10 ^W/m2

'Phdn tich m huong ddn gidi

W M u c cuong dp am tai A va B Ian lupt la: = l O l o g — va L g = l O l o g —

Hieu muc cuong dp am tai A va B la:

Vl d u 8 : Ba diem O, A, B cimg nam tren mot nua duong thSng xuat phat tiJ' O

Tai O dat mot nguon diem phat song am d i n g huong ra khong gian, moi truong khong hap thu am Muc cuong dp am tai A la 100 dB, tai B la 60 dB

Muc cuong dp am tai diem M thupc doan AB voi A M = 4 M B la _ A 61,9dB B 72,6dB C 43,6 dB D 70,5 dB

^hdn tick vd hu&ng ddn gidi

V i d u 9 : M o t dan loa phat am thanh dSng huong M u c cuong dp am do

dupe tai cac diem each loa mot khoang a va 5a Ian lupt la lOOdB va L Gia

trj ciia L la

A lOOdB B 4 9 d B C 86dB D 25dB

'Phdn tich vd hic&ng ddn gidi Gpi m u c cuong dp am tai diem each loa m o t khoang a la L^

Gpi m u c cuong dp am tai diem each loa mot khoang 5a la L^^

H i ^ u muc cuong dp am tai hai diem do la:

L, -L53 = l O l o g i ^ = l O l o g f - 10.1og25 = 14

= L a- 1 4 = 1 0 0 - 1 4 = 86dB

C h g n dap an C

V l d u 1 0 : (Trich de thi thu chuyen Ha T i n h Ian 1 nam 2013)

Xet d i e m M 6 trong m o i t r u o n g dan hoi c6 song am truyen qua M i i c

cuong d p am tai M la L (B) Ne'u cuong dp am tai diem M tang len 100

Ian thi mue cuong dp am tai diem do bang

Chii y: mi'fc cuang do dm trong cong thiec tren tinh theo dan vi deciben (dB)

nhimg decho muc citang do am tai M la Ben (B) vi the can doi ra dB

L(B) = 1 OL(dB) nhu vay dap an A chua dung De dang thay ring dap an phdi Id

B Vai nhung cau the nay, ne'u cdc ban khong dey den dan vi thi rat de "mac bay" ciianguairade

Chpn dap an B

Vl d u 1 1 : (Trich de thi thu chuyen Ha T i n h Ian 2 nam 2013)

M o t n g u o i d u n g each nguon am mot khoang la x t h i cuong dp am la I

K h i n g u o i do tien ra nguon xa them mpt doan 30m t h i cuong dp am

giam chi con bang 1/4 Khoang each x ban dau la

A 60m B 15m C 7,5m D 30m

'Phdn tich vd hu&ng ddn gidi

Gpi A la d i e m each nguon am mpt khoang x (x > 0) c6 cuang dp am 1^ = I Gpi B la d i e m each nguon am mpt khoang x +30 c6 cuong dp am Ig = -

4

Ta biet rang cuong dp am ti

le nghjch v o i b i n h p h u o n g khoang each t u diem can tinh cuong dp am den nguon nen

V i d u 1 2 : Tai O c6 1 nguon p h a f a m thanh dSng h u o n g v d i cong suat

khong d o i M p t n g u o i d i bp t u A den B theo 1 d u d n g thJing va ling nghe

, am thanh t u nguon O thi nghe thay cuong dp am tang t u 41 den 251 roi

iji i/ui/i'i uji miivn nil uiii iw

(Phdn tich vd huang ddn gidi

Theo bai ra ta c6: cuong do am tai A va B c6 gia trj bang nhau ket hop v6i

gia thie't de cho la song am truyen duoi dang song cau (am phat dSrig

huong) vi the A va B ciing nam tren mot duong tron Vay diem c6 cuong do

am Ion nhat tren doan A B chinh la diem H nam tren duong trung true cua

tam giac O A B (nhu hinh ve)

Theo bai ra: 1 ^ = I B = 41 => O A = O B

V I du 13: (Trich de thi t h u chuyen Ha L p n g - Quang N i n h Ian 1 nam 2013)

Tai hai diem P va Q trong khong khi c6 hai nguon song am cung tan so f,

cijng bien do A , do Ifch pha la TI Song am truyen t u hai nguon am do voi

buoc song A den diem N nam ngoai duong thang PQ, c6 hieu khoang

each den P,Q la k A ( v o i k = 1,2,3 ) Coi moi truong khong hap thu am

Khi do, tai diem N

A hai song giao thoa nhau ung voi bien dg cue tieu la A N = 0

B hai song giao thoa nhau ung voi bien do cue dai la A N = 2A

C hai song giao thoa nhau ung voi bien do la A N = A 7 2

D hai song khong giao thoa nhau nhung c6 bien dp song la A N =A 0

'Phdn tich vd huong ddn gidi

Hai song tren thoa man dieu kien giao thoa vi the m u o h biet tai mot diem

dao dong voi bien dp cue dai hay cue tieu ta lam n h u sau:

A p d u n g cong thuc tinh tong quat ben giao thoa:

27t,

d , - d i = ( A < P N + A ( p ) — = > A ( P N = - - ( d 2 - d i ) - A ( p

^ In K

Theo bai ra: d 2 - dj = kX va A(p = rt

Thay vao: A c p ^ = — " ( ^ 2 ) - Acp = - ^ k ) - n = (2k - 1 ) 7 1 thuoc so le Ian TT

K A

nen tai N song dao dpng voi bien do cue tieu Vi bien dp cua hai nguon bang nhau nen bien dp tai diem N bang 0

Chon dap an A

Vi du 14: (Trich de thi thu chuyen nguyen Trai - Hai Duong Ian 1 nam 2013)

Mot am thoa c6 tan so 440 Hz (phat ra am la) dat sat mieng mot binh tru dung nuoc c6 muc nuoc each mieng binh sao cho am thanh phat ra t u mieng binh la to nhat H o i can rot them vao binh mot cot nuoc c6 chieu cao toi thieu la bao nhieu thi am thanh tro nen nho nhat? Van toe truyen

am trong khong khi bang 330m/s

A 18,75cm B 17,85 cm C 37,5 cm D 27,5 cm

^hdn tich vd huang ddn gidi

Song d u n g tao ra trong binh hinh tru mot dau kin, mot dau ho c6 chieu dai binh thoa man:

ra am to nhat ung voi hai gia trj tan so cua hai hpa am lien tie'p la 150 Hz

va 250 Hz Tan so am nho nhat khi ong sao phat ra am to nhat bang

A 50 Hz B.75HZ C 25 Hz D 100 Hz

'Phdn tich vd huang ddn gidi

Chieu dai 6'ng sao thoa man:

U k ^ 4 = ( 2 k l ) ^ = ( 2 k l ) ^ ^ f = ( 2 k l ) ^ = > ^ = ^

Bi quyet on luy?n thi dai hgc dat diem tot da Vat It, tap 1 - Le Van Vinh

Vi du 16: (Trich de thi dai hpc 2012) |

Tai diem O trong moi truong d i n g huong, khong hap thu am, c6 2 nguon

am, giong nhau v a i cong suat phat am khong doi Tai diem A c6 miic cuong

do am 20dB De tai trung diem M cua doan O A c6 miic cuong do am la

30dB thi so'nguon am giong cac nguon am tren can dat them tai O bang

A 4 B 3 C.5 D 7

'Phdn tick vd hu&ng dan gidi

M u c cuong do am tai M k h i tai O c6 hai nguon am:

I

L M - L A = 1 0 1 o g - ^ = 101og O A

= 10 log 20M

[OM OM)

Goi IM , L M la cuong do am va muc cuong dg am tai M k h i tai O c6 2 nguon diein

Goi I ' M , L ' M la c u o n g do am va m u c cuong do am tai M k h i tai O c6 n

=> Tai O CO tat ca 5 nguon am nen suy ra can dat them 5 - 2 = 3 n g u o n a m

n t r a de tai M c6 muc cuong do am la 30dB Chgn B

Vl du 17: (Trich de thi thu chuyen nguyen Trai - Hai Duong Ian 1 nam 2013)

Trong mot buoi hoa nhac, k h i d u n g 10 chiec ken dong t h i tai cho ciia mot

khan gia do dugc muc cuong do am 50dB H o i phai diang bao nhieu

chiec ken d o n g de tai cho khan gia do c6 muc cuong dp am la 60dB?

A 50 B 80 C.lOO D 90

'Phdn tich ra huorng ddn gidi

Goi I la cuong dp am do mot chiec ken phat ra tai v i tri do muc cuong dp am

C u o n g dp am do 10 chiec ken phat ra tai v i tri do se la 101 Gpi n la so ken de tai v i t r i tren do dupe muc cuong dp am la 60dB v i t h e t a c6:

Vi du 18: Cong suat am thanh cue dai cua m p t may nghe nhac gia d i n h la

l o w Cho rang c u truyen tren khoang each 2m, nang l u p n g a m b i giam 6% so v o i Ian dau do su hap t h u cua m o i t r u o n g truyen am

Biet lo = 1 0 - ' 2 W / m 2 N e u m o to het muc t h i m u c cuong dp am 6 khoang each 6m la:

A 102,6 dB B 107dB C 99 dB D 88 dB

'Phdn tich vd hitong ddn gidi

Day la bdi loan md trong qud trinh truyen am c6 su hap thu dm Vi co su hap thu

am nen nang luang dm gidm dan trong qud trtnh truyen Degidi dugc dang bdi todn ndy chung ta phdi ndm dugc moi lien he giita nang luang dm vd cong suat dm: E = Pt

Theo bai ra: C u 2 m t h i nang l u p n g am giam 6% so v o i nang l u p n g ban dau, khi do ta co:

Sau 2m dau tien: = 0,06 ^ ^ = 0,94 Sau 2m tiep theo (4m): =>-?- = (0,94)^

Sau 2m tiep theo (6m): => = ( 0 , 9 4 f Cong suat am tai v j t r i each nguon am 6m la:

Pnt = ( 0 , 9 4 ) ^ = > P 3 = ( 0 , 9 4 f Po

P Pn.(0,94)^

C u o n g dp am phat d i t u nguon diem dupe xae d i n h la: I3 = -3- = - '2—

S 47td

Vay muc cuong cto am tai v i tri each nguon am 6m la:

Chon dap an A ' '

ta B A I T A P V A N D U N G

C a u 1 : (Trich de t h i thtr chuyen dai hQC V i n h Ian 3 nam 2012) ^

A m do mot chie'c dan bau phat ra

A nghe cang traim khi bien do am cang nho va tan so am cang Ion

B nghe cang cao khi ipuc cuong do am cang Ion

C CO do cao phu thuoc vao hinh dang va kich thuoc hop cong huong

D CO am s^c phu thuoc vao dang do thj dao dong cua am

'Phdn tick vd hic&ng ddn gidi

A m do mot chie'c dan bau phat ra

A (sai) v i am nghe cang tram khi bien do am cang nho va tan so am cang

l e « nho

B (sai) v i am nghe cang cao khi tan so am cang Ion khong phai muc cuong

dp am cang Ion

C (sai) vi am ccS do cao phu thuoc vao tan so

D (diing) vi am ScIc la mot dac trung sinh ly ciia am va p h u thuoc vao dang

do thi dao dong ciia am

Chpn dap an D

C a u 2 : Cuong do am chuan la IQ = 1 0 " ' ^ W / m ^ Cuong do am tai mot diem

trong moi truong truyen am la lO^^W/m^ M u c cuong am tai diem do bang

c a u 4 : (Trich de thi t h u chuyen nguyen Trai - Hai D u o n g Ian 1 nam 2013)

A m do hai nhac cu khac nhau phat ra luon khac nhau ve:

A Dp cao B Do to C A m s5c D Ca A, B va C

'Phdn tich vd hii&ng ddn gidi

Cau nay tha'y ngan the nay nhung hau het cac ban deu boi roi vi trong dap

an CO dap an D la dap an dung cho ca 3 dap an con lai Chu y rSng: am ScIc gan lien voi do thj dao dong vi the cac nhac cu khac nhau khong the c6 Cling do thi dao dong vi the khong the cimg am sac dugc nen chi co dap an

C dung

Chpn dap an C

C a u 5 : (Trich de t h i t h u T H P T N h u Thanh - Thanh Hoa Ian 2 nam 2013)

Muc cuong do am L cua mot am c6 cuong do am 1 dugc xac dinh being cong thuc (In la cuong do am chuan)

Phdn tich vd huang ddn gidi

Muc cuong do am dugc xac djnh theo cong thuc: L(dB) = l O l g y Chpn dap an A

C a u 6 : (Trich de t h i t h u chuyen Dai Hpc V i n h Ian 1 nam 2013)

Cuong do am tai diem A each mot nguon am diem mot khoang I m b e i n g

10 *'W/m'^ Cuong do am chuan b^ng 10 '-^VV/m^ Cho rang nguon am la nguon dang huong va moi truong khong hap thu am Khoang each t u nguon am den diem ma tai do muc cuong do am b5ng 0 la

A 750m B 250m C 500m D 1000m

Phdn tich vd huang ddn gidi * •

Ggi B la diem c6 muc cuong do am bang 0 Theo bai ra ta c6: :

L A - L B = 2 0 1 o g Chpn dap an D

R = 60 => log

A J

= 3 = ^ R B = 10-^RA = 1000m

C a u 7 : (Trich de thi thu T H P T T h i ? u Hoa - Thanh Hoa Ian 2 nam 2013)

Phat bieu nao sau day khong dung?

A D o cao ciia am la mot dac ti'nh sinh If ciia am

B Nhac am la do nhieu nhac cu phat ra

C Tap am la cac am c6 tan so'khong xac d i n h

D A m sac la m^t dac t i n h sinh l i ciia am

'Phdn tick vd huang ddn gidi

A (dung) v i do cao cua am la mot dac tinh sinh l i ciia am gan lien v o i tan so

am

B (sai) V I nhac am la do m p t hay nhieu nhac cu phat ra deu dugc

C (diing) v i theo d i n h nghla: Tap am la cac am c6 tan so k h o n g xac dinh

D (dung) vi am sic la m p t dac tinh sinh li cua am gan lien v o i d o thj dao

dong am

ChQn dap an B

C a u 8 : Ba d i e m O, A , B cung nam tren mot nua d u o n g thang xua't phat tu O

Tai O dat mot nguon diem phat song am dang h u o n g ra khong gian, m o i

t r u o n g khong hap thu am M i i c cuong dp am tai A la 70 dB, tai B la 30 dB

M i i c cuong d o am tai trung diem M ciia doan A B la

A 4 5 d B B 5 4 d B C 36 dB D 75 dB

^hdn tick v>d huomg dan gidi

Theo bai ra, ta c6:

C a u 9 : H a i diem A va B nam 6 cung 1 phia ciia nguon am, tren ciing 1 phuonj^

truyen am c6 m i i c cuong dp am tai A va B Ian lugt la: L A = 80 dB; LB = 20 ciP

Neu nguon am do dat tai A thi muc cuong do am tai B k h i do la

onyi UIB uonSu UBp B X ^ q ; oaq; 'Sueq §ugq; j 'g 'y uiaip e o q ^ -.jj nej

a U B dep u 6 q 3 aPOOl = 03 + 08 = 03 + "^1 = ^1 <=

" J O I U I B uonSu qDBD uiaip 4 0 U I I B X " J ? ex UBqd B A riqi dsq x\s 00 8uoq>i ris

B I 3 uBig Suoq>f guoj4 Suonq §uBp U I B 4Bqd uiaip uonSu G\B uon§u ]6y^

(£103 ui?u £ u?i q u j N D ? g u3Xnq3 i q ; ap q D u ^ ) : 0 T nej

:v\u Drq g IB4 U I B 6p SuoriD iqi y 1B4 4 B p uie uonSu i\\y{

va666= ^ a - 3 a = av^^aoooi=3a 9 a = oOl =

a ' I

09 = 03 - 08 = vjSoioi = ^ 1 - ^ 1 B4 'BJ iBq o a q i

iviS uyp Sujpnif ya uyii^

'Phdn tich vd huong ddn gidi ^ , j

Gpi P va I la cong sua't va cuang dp am aia moi nguon

P P Khi do ta c6: I = — = r

S 4nR^

Gpi IM ,LM la cuong dp am va muc cuong dp am tai M khi tai O c61 nguon die'm

Gpi I'M ,L'M la cuong dp am va muc cuang dpamtaiMkhitaiOcon nguon diem

L M - L M = 1 0 1 o g ^ = 6 0 - 4 0 = 2 0 = ^ ^ = 20 nPi

Tgi O CO tat ca 20 nguon am Chpn dap an A

Cau 15: Mpt nguoi dung giua hai loa A va B Khi loa A bat thi nguoi do

nghe dupe am c6 muc cuong dp 60dB Khi loa B bat thi nghe dupe am c6

muc cuong dp 70 dB Neu bat ca hai loa thi nghe dupe am c6 muc cuang dp

bao nhieu?

A 130dB B 70,4dB C 60dB D 70dB

'Phdn tich vd huang ddn gidi

Gpi Ij va I2 la cuang dp am loa A va loa B tai diem do Khi do cuang dp

am tai diem do khi bac ca hai loa la: I = Ij +12

Theo bai ra, ta c6:

Muc cuang dp am cua cua loa A: Lj = lO.log— = 60 => Ii = IO^IQ

Muc cuong dp am cua loa B: = lO.log 1?- = 70 => Ij = 10^IQ

Miic cuong dp am khi bac ca loa A va loa B:

L,2 = 1 0 1 o g i l ^ l 0 1 a g l ^ i l I ^ = 70,4dB

Chpn dap an B

Cau 16: Hai hpa am lien tiep do mpt day dan phat ra c6 tan so han kem nhau

la 56Hz Hpa am thu 3 c6 tan so la

A.28HZ B.56HZ C 84Hz D 168Hz

i'hdn tich vd hucmg ddn gidi

Theo de ra, ta c6 nf - (n -1)f = 56 => tan so am ca ban f = 56Hz

Tan so hpa am thu 3: £3 = 3f = 3.56 = 168Hz Chpn dap an D

^hdn tich vd hudng ddn gidi

Gpi Ij va I2 la cuang dp am tai va cuang dp am phan x^ tai diem do Khi

do cuong dp am toan phan la: I = Ij +12 M Theo bai ra, taco :

Muc cuang dp am ciia am truyen toi: Lj = lO.log— = 10 => Ij = IOIQ Muc cuong dp am cua am phan xa: L2 = lO.log — = 20 => I2 = IOOIQ Muc cuong dp am toan phan:

L,2 =10.1oglilil = 10.1ogl^^5^ti^==20,4dB

Chpn dap an D

CSu 13: Trong mpt ban hpp ca, coi mpi ca si deu hat vai cung cuong dp am

va cung tan so' Khi mpt ca si hat thi muc cuang dp am la 60dB Khi ca ban hpp ca cung hat thi do dupe muc cuong dp am la 90dB So ca si c6 trong ban hpp ca la

A 10000 nguoi B 100 nguoi C 10 nguoi

Phdn tich vd hit&ng ddn gidi

Gpi so' ea si la n, cuang dp am cua moi ca si la I Suy ra cuong dp am eiia n ea si se la: = nl

60 dB thi tai O ta phai dat tong so nguon am giong nhau la

A n = 20 nguon B n = 50 nguon

C n = 10 nguon D n = 100 nguon S

Cty TNHH MTV DWH Khang Viet

Cau 19: Song (A, B cung phia so voi S va A B = 100m) Diem M la t r u n g dieVn

A B va each S 70 m c6 muc cuong dp am 40dB Biet van to'c am trong khong

k h i la 340m/s cho r^ng moi t r u o n g khong hap thu am (cuong dp am chuaVi

Ii, = lO'^W/m^) Nang lupng ciia song am trong khong gian gioi han boi hai

mat cau tarn S qua A va B la

A 207,9nJ B 207,9 m j C 2a7mJ D 2,07^J

'Phdn tich v>d huong dan gidi

Song truyen trong khong gian Nang l u p n g song ti I f nghjch v o i binh

p h u o n g khoang each Nang l u p n g song b i n g gi? 6 day de y cho muc

cuong d p am tai diem M la trung diem AB, nghla la se xac d j n h dupe cuong

dp am tai M Can cu suy ra cuang dp am tai A va B C u o n g dp am tai A va

B ti le nghjch voi binh p h u o n g khoang each d o n v j la W/m ^ N a n g lupng

song tai cac mat cau tam (S, SA) va (S, SB) Lay h i f u t h i dupe nang lupng

Cuong do am tai 1 diem la nang l u p n g d i qua mpt d a n vj d i f n tich tinh

trong 1 d o n vj thoi gian T u gia thiet suy ra cong suat nguon S la

„uye't on luyftt thi dai hqc dat diem toi da Vat U, tap 1 - Le Van Vinh

Cau 17: M o t 6'ng khi c6 mpt dau bjt kin, mpt dau h o tao ra am co ban c6 tan

so n 2 H z Biet toe dp truyen am trong khong k h i la 336m/s Buoc song dai nha't cua cac hpa am ma ong nay tao ra b i n g :

Cau 18: Tai mot diem tren mat phang chat long co mpt nguon dao d p n g tao

ra scSng on d i n h tren mat chat long Coi m o i t r u o n g tuyet doi dan hoi M va

N la 2 diem tren m|it chat long, each nguon Ian lupt la R i va R 2 Biet bien dp

dao dpng cua phan t u tai M gap 9 Ian tai N Ti so ^ ^ a n g

'Phdn tich va hu&ng dan gidi:

N a n g l u p n g song co ti le voi binh phuong bien dp, tai mpt d i e m tren mat phang chat long co mpt nguon dao dpng tao ra song on d j n h tren mat chat long thi nang l u p n g song truyen di se dupe phan bo deu cho d u d n g tron (tam tai nguon scSng) Cong suat t u nguon truyen den cho 1 d o n vj dai vong tron tam O ban k i n h R la

R, 81

Cty TNHH MTV DWH Khang Vift

Bieu thuc hi^u di^n thehai dau mach:

u = Uo cos((ot + (Pu) = 2Q0\[l cos

Bieu thuc hieu di?n the hai dau d i f n tro R:

U R = UoRCOs(cot + (PuR ) ; Voi: UOR= IO.R = 2.100 = 200 V ;

Trong dean mach chi chua R thi u ^ cung pha i :

Vay: U R = UORCOS (cot + cPu^ ) = 200cos lOGTit V

Bieu thuc hi^u di^n the hai dau cupn cam L:

U L = UoLCOs(cot + cpuj^) V a i : UOL= IO.ZL = 2.200 = 400 V;

Trong doan m^ch chi chua L thi U L nhanh pha hon cuong dp dong di?n ^

71 - 7t 7t ,

<PuL = ^ i + 2 " ° ^ 2 " 2

Bieu thuc hi^u di^n the hai dau ty di^n C:

U c = Uoccos(cot + (Pu^) V o i : Uoc= lo.Zc = 2.100 = 200V;

Trong doan mach chi chua C thi xx^ cham pha hon cuong dp dong d i f n ^

71 7t

^ u c = < P i - f = 0 - ^ = - ^ rad

Vay: uc= Uoccos(cot + (p^,„) = 200cosflOOTtt- -1V

2

Vi dy 2 (Trich de thi dai hpc 2013):

Dat di^n ap u = 22o72 cosl007it (V) vao hai dau doan mach mkc nol tiep

Bieu thuc u hoac i se luon c6 dang:

u = Uo.cos(cot + cpy) hoac: i = Io.cos(cot + cpi)

Vi vay de viet dupe bieu thuc cua hieu dien the hoac cuong dp dong di$n chiing ta can phai xac dinh 4 yeu to' la Uo, lo, co va cp Sau do diing cong thuc: cp = cpu - cpi

Voi cp la dp l^ch pha giua hi^u dien the so voi cuong dp dong di^n f

cp dupe xac dinh theo cong thuc: tgcp = — — ^

Lttu y: Mach khuyet phan t u gi thi trong cong thuc tren, ta khong dua vao

Cdc trudng hop dac biet: ^

+ Doan mach chi chua di^n tro thuan R thi cp = 0

+ Doan mach chi chua cupn thuan thi cp = +—

Cam khang: = L.co = -IOO71 = 200Q

Pha ban dau cua hifu di^n the: (p, = (p^ cp = 0

-Bieu thiic cuong dp dong dien: i = I„ cos(ci)t + (pj) = 2cos

Chpn dap an C

= — rad

100TCt +

-4j

V i d u 4 : Dat mot dien ap xoay chieu on djnh vao hai dau doan mach gom

mpt di^n tro R = 50Q va mgt cugn day thuan cam c6 L = j - (H) thi dong

dien qua mach c6 dang i = 2cos(1007Tt - ) (A) Neu thay dien tro R bang

tu dien C thi cuong do hieu dung qua mach tang len y/l Ian Bieu thuc i

cua dong dien qua mach sau khi thay la

A i = 2V2 cos(100Tit- ^ ) (A)

C i = 2V2cos(1007Tt - — )(A)

B i = 2V2cos(1007Tt+ -)(A) 4

^hdn tich vd humig dan gidi

Cam khang: Z L = wL = IOOTI.— = 50Q

Chpn dap an B 1007rt + -4

Vl d u 5: Neu dat vao hai dau mot mach dien chua mot dien tro thuan va

"^Qt cupn cam thuan mgc noi tiep mgt di^n ap xoay chieu c6 bieu thuc

<Phdn tich v>d huang dan gidi:

Cam khang : = L.co = - I O O 7 : = lOOQ;

Dung khang : ZQ = «.C

IOOTI 10 1-4 = 200Q 27:

- To'ng tro : Z = + (Z^ -Z ^f = JlOO^ + (100 - 200)^ = \m42Q

Vl d u 3 : Cho mach dien AB, trong do C = ;^10 "F, L = r ^ H , R = 25n 37:"' ' 271' _1_

mic noi tiep Bieu thiic di^n ap giiia hai dau mach U^B = 50 %/2 cos 1007:tV

Viet bieu thuc cuong dp dong dien trong mach ?

A i = 2cos

C i = 2cos

1007lt 4 1007lt + -

'Phdn tich vd huang dan gidi

Cam khang: Z L = L.co = — I O O T I = 50Q

Ctu I Mil I MIVDWHKhang Viit

Vl d u 8 : Cho ba linh ki^n R = 60Q, cupn day thuan cam L, tu di^n C Lan

luot dat dien ap xoay chieu c6 gia tri hi?u dung U va tan so f khong doi

vao hai dau doan mach noi tiep RL hoac R C thi dong di^n qua mach c6

cac bieu thuc ii = V2 cos(100nt - ) (A) va i2 = V2 cos(100TTt + ^ ) (A)

Neu dat di^n ap tren vao doan mach RLC noi tiep thi dong di^n qua

mach CO bieu thuc:

A.i = 2cos(100nt+ -)(A)

(Phdn tick v>d hudng dan gidi

Theo bai ra ta c6: IQI = I02 => Zi = Z2 \I^^+Zl = JR^TZJ => Z L =

tan(pi

-tan(p2 = R

Z L - Z C

• tancp] = -tan(p2 =>(Pi = -92

Trong do: (p,; (p2 la do l^ch pha ciia di^n ap 2 dau mach voi i^; ij

_n 7n

Tudo (D, =0),, -cpi, =^1 ^'1 4 — I -f-—1 I2J 3 3 = —rad=>(P2 = - ^ r a d

tancpi = % = ^/3 ZL = 60>/3Q R

Vi ZL; Z(~ la cac h3ng so ma Z ^ = ZQ nen khi dat di^n ap tren vao mach

RLC noi tiep thi hi^n tugng cgng huong xay ra nen Z = R va u cung pha voi i

Do do cuang dp dong di^n eye dai trong mach:

Bi quyet on luy$n thi dai hQc dat diem toi da Vat It, tap I - Le Van Vinh

u = 100N/2 cos(wt + -) iV), thi khi do di|n ap hai dau di#n tra thuan c6 4

bieu thuc UR=100cos(o)t) (V) Bieu thuc di#n ap giua hai dau cupn cam thuan se la

A. U L = 100cos(wt+ - ) ( V )

C U L =100cos(cot+ ^)(V)

B. UL =100N/2cos(cot+ ^ ) ( V )

D. UL =100V2 cos(cot+ ^ ) ( V )

^hdn tick m hu&ng ddn gidi

Su dung may tinh:

Taco: u = UR + UL = ^ UL = u - UR = 100x/2Z J - lOOZO = lOOZ^

Vay: u =100cos Chgn dap an A

<Phdn tick ra hu&ng ddn gidi

Vi mach chi c6 ty nen: (Pi==(pu+r = 0 + X = T ^0=^ = COCUQ

(A) (A)

(phdn tich ra huong ddn gidi

Hi^u di?n thehi^u dung giira hai dau di?n tro:

CUf TNHH MTV nwil Khaug Viet

^hdn tich vd hit&ng dan gidi

Thong thuang khi vic't bic'n thi'ic dien dp ci'ia hai dau doan mach ndo do ta phdi tinh

duox dien dp cite dai gifta hai dau doan mach do vd do lech pha cua no so vai cwdii'^

do dong dien trong mach Doi voi bni nay tht de Idm nhw vqy rat ddi so v&i tiiai

gian cua mot bdi trdc nghiem

Tuy nhien ta c6 the Idm vice do tuong doi dan gidn neu chii y den nhung giu kien

bdi todn cho

Theo gia thiet \Jc = U K ^ = R

Gia trj ciia L de Ui eye dai thoa man: Z[ = — - — — -> Z| = 2R

D Q l^ch pha g i u a UKi va i : tancp,., =-?L = 2 ^ ( n n , = — \

R 1 8 0 '

Khi Ui cue dai , thi U I m., = -^-j^ ^ - UN/2 = IOOV2 ( V )

Ket hop v d i bieu thuc - Uf< + ( U L - U C ) ^ U K = ^ = 50V2V

U R L = = ^ ( 5 0 7 2 ) ' + ( 1 0 0 / 2 ) ' = 50V1OV

Vi U I max nen U ^ c 1 U > ip (pK(^ =

Z Mat khac tancp^;^ =^ — ^ = - i ^ =

Vl du H : Dat mot dien ap xoay chieu u = 200 cos(100TTt + ^) vao hai dau

doan mach m^ic nol tiep theo t h u t u cac phan t u gom m o t dien t r o thuan

R, m o t tu di$n eo dien d u n g C va mot cupn cam co do t u cam L thay d o i ,

duoc Dieu chinh L de dien ap gii>a hai dau cuon day dat cue dai t h i k h i

do di$n ap giCra hai dau cua tu dien la Uc = 100 V Hay viet bieu thiic

giOa hai dau ciia R va C

Bi quyct on luyftt thi dai hqc dat diem toi da Viit I"!' ' - Van Vinh

j(V) vao hai dau m p t t u d i ^ n c6

d i ^ n d u n g ^'^^ (F) 6 thoi diem di^n ap giua hai dau tu dien la 150V

V l du 1 0 : Dat vao hai dau doan mach m p t d i f n ap xoay chieu

u - 100N/2cos(a)t) (V) gom mpt di?n tro thuan R, cuon day thuan cam c6

d p t u cam L thay d o i dupe va mpt tu d i ^ n c6 dien d u n g C K h i thay doi

L ta thay Ui dat cue dai va h i ^ u d i ^ n the hai dau t u d i ? n b^ng hi?u di?n

t h e h a i dau d i ? n t r o thuan Viet bieu thuc di?n ap giija hai d a u cua U ^ L

Pha ban dau ciia hi#u di#n the: (p = (Pu-9i=>'Pu =<Pi+<p=-— + — = 0

4 4 Bieu thuc hieu dien the: u = UQ cos(cot+ (pu) = IOOV2 cos(1007it)V

Chpn dap an D

Cau 2: Cho mach di?n gom RLC noi tiep

Di^n ap hai dau mgchu = 220^2 cos(1007tt + ^ ) ( V ) Di?n tro R = 50>/3 L

la cuon day thuan cam c6 L = — H , di^n dung C = ——F, bieu thuc cuong

dp dong di^n trong mach:

'Phdn tick vd hu&ng ddn gidi

Cam khang: = L.to = -^1007: = 50Q

C a u 3 : D^it di^n ap xoay chieu u = U o C O s lOOntn (V) vao hai dau mpt

cupn cam thuan c6 dp tu cam L = —(H) 6 thoi diem dif n ap giiia hai dau

cupn cam la 100 N/2 V thi cuong dp dong di^n qua cupn cam la 2A Bieu

thuc cua cuong dp dong di?n qua cupn cam la

'Phdn tich vd hu&ng ddn gidi "

Theo gia thiet khi L thay doi Ui dat cue dai thi:

U R C vuong goc voi UAB : U L M A X ^ = ^RC^ + \ J ^ = U^^ + VQ^ +

Mat khac: = UR^ + (ULMAX - Uc)' Voi U = 100>/2V; Uc = lOOV => UL = 200V

Ta CO U R C ' = U L M A X ' ' ^ ^ ^ URC - IOOV2V va cpRc = | - ^ = -^(rad) Bieu thuc di^n ap hai dau URC = 200 cos(1007it - —)

<Phdn tich vd hu&ng <Mn gidi

Cam khang: = L.co = -IOOT: = lOOn;

Tong tro: Z = ^ M ^ T ^ = /so^+(100-50)^ = 50V2Q

Hieu di?n the cvrc dai: Uo = I0.Z = 2.50V2 V =100 ^ V

Dp l?ch pha: tancp = - Z L ^ Z C ^100-50 , 7t

R 50 = l=>(p = —rad 4

Z83

n , r ( A ) - + 11^001

V UB dep u6n3 soDQt' = n i q D B u i n e p l e q n o r i q } n a i q Xey\

(phdn tick vd hicdng dan giai

Cam khang: Z L = o)L = IOOTT.— = 50Q Mach chi c6 cuon cam thuan nen i tre pha hon u mot goc ^ :

Cau 4 : Khi dat hieu dien the khong doi 30V vao hai dau doan mach gom dien tro thuan mac noi tiep voi cuon cam thuan c6 do tu cam — (H) thi dong

471

dien trong doan mach la dong dien mot chieu co cuong dq> l A Neu dat vao

hai dau doan mach nay dien ap u = 150N/2 cosl207t t (V) thi bieu thuc cua cuong do dong dien trong doan mach la

471

Khi dat hi§u dien the khong doi vao hai dau mach thi chi c6 R gay can tro

con Zl = 0 R = _ U|c _ 30 = 30Q

lie 1 Khi dat hiC^u dien the xoay chieu vao hai dau mach thi c6 ca R va ZL gay can

tro nen tong tro cua mach: Z = + Zl = 730^+30^ - SOsflQ

Cuong do dong dien cue dai' IQ = = = 5(A)

Dp l?ch pha giua u va 1: tancp = ^ S Q ^ T

Co mot each tinh cue nhanh ma cac ban da duoc lam quen trong cuo'n 'ca'm

nang luy#n t h i dai HQC vat ly tap 1' la tinh theo may tfnh casio fx 570 ^u,

hoac casio fx 570 es plus

Hi?u d i ^ n the tuc thoi hai dau mach : u = i.Z = — Z

Bam may: ^.lOx/2 = 4 0 Z

-^ 10 4

Vay: u =40 cos 1007rt +

-4 j (V)

Cau 6: M 6 t doan mach dien xoay chieu gom mot dien tro thuan R = 80Q, mot

cuon day thuan cam c6 do t u cam L = 64mH va mot tu dien c6 dien dung

C = 40nF mac noi tiep Biet tan so cua dong dien f = 50Hz Doan mach duoc

dat vao dien ap xoay chieu co bieu thuc u = 282cos314t (V) Lap bieu thiic

cuong do tuc thoi cua dong dien trong doan mach

^hdn tich m huang ddn gidi

Tansogoc: co = 2Trf = In.SO = lOOn rad/s

Cam khang: Z L = coL = 10071.64.10"^ « 20Q

Cau 7: M o t doan mach gom cuon cam c6 do t u cam L va dien tro thuan r mac

noi tiep voi tu dien c6 dien dung C thay doi duoc Dat vao hai dau mach

mot hieu dien the xoay chieu c6 gia trj hieu dung U va tan s o f khong doi

Khi dieu chinh de dien dung cua tu d i ^ n c6 gia trj C = C i thi dien ap hieu

dung giOa hai dau tu dien va hai dau cupn cam c6 cung gia trj va bang U,

cuong do dong dien trong mach khi do c6 bieu thuc ij =2V6co6 1007rt+- (A)

Khi dieu chinh de dien dung cua tu dien c6 gia trj C = C 2 thi dien ap hieu

dung giua hai ban tu d i f n dat gia tri cvrc dai Cuong dp dong d i ^ n tuc thoi trong mach khi do c6 bieu thuc la

A i2=2V2cos(1007rt + 57t/12)(A) B i2 = 2^cos(1007tt + 7t/3)(A)

C i2=2V3cos(1007rt + 57r/12)(A) D i2 = 2^/3cos(1007tt + 7t/2)(A)

<Phdn tich v>d huang ddn gidi

Khi C = Ci => U L = U c = U (cong huong) => U R = U = U L = U c ^ R = Z L

Cau 8: Dat dien ap u = 100 N/2 COS 1 0 0 7 t t

-2 (V) vao hai dau mot doan mach

25,._2, gom mot cuon cam c6 dien tro thuan r = 5Q va do t u cam L = —10 H mac noi tiep voi mot d i f n tro thuan R = 20Q Bieu thuc cuong do dong di?n trong doan mach la

A i = 2^/2 cos(1007Tt - n/4 ) (A) B i = 4 cos(1007it + 7i/4) (A)

C i = 4cos(1007it-37r/4)(A) D i = 2N/2 cos(1007rt + 7i/4) (A)

^hdn tich vd huang ddn gidi

It 71 3n

Ma (p = cpu - (Pi ^ cPi = (Pu - ^ = ^2 ~4

Cuang do cue dai: IQ = 7(R + rf + Zl 725^+252 Un 100^/2 = 4A

Vay bieu thuc cuong do dong di^n trong truong hg-p C = :

3TV

i = 4cos

100-Kt Chpn dap an C

Cau 9: Dat dien ap xoay chieu u = 100>/2 coscot (V), co thay doi dugc dat

vao hai dau doan mach AB gom hai doan mach AM va MB mac noi tiep

Doan mach AM gom bien tro mac noi tiep vol cupn cam thuan, doan mach

MB chi CO mot tu dien Khi co - 1007i(rad/s) thi dien ap hieu dungU^,^,

khong phu thuoc vao gia tri cua bien tro, dong thai dien ap hi^u dung

1007rt + -3 ( V )

6 (V)

^hdn tich m huang dan gidi

Hieu dien the hieu dung hai dau AM:

Dp Ipch pha giiJa u va 1: tancp =

Ma (p = (pu - (Pi =^ (Pi = cp^, - (p = 0

Cty TNHH MTV nV\lI Khtmg Viet

Ma <PAM ='PUAM ^

Vay: UAM =100N/2COS

71 TT T:

6 6 1007lt + - V

Chpn dap an A « ' >

CaU 10: Doan mach dien xoay chieu AMB cau tao gom doan AM chua R va

C mac noi tiep voi doan MB chua cupn cam thuan c6 L thay doi Dien ap xoay chieu hai dau mach AB: u = 7572cos( lOOTit+^] (V) Dieu chinh L den

'Phdn tich vd huang dan gidi

L bien thien de Ui max nen

'Phdn tich vd hic&ng dan gidi

Bai nay giai theo may tinh se cue ky nhanh:

u = i.Z = V2Z0.(10 + 10i-20i) = 2 0 Z - - " '

4 &

Vay: u = 20cos Chpn dap an A

1007rt lv

Bi quySt on luyftt thi aai hqc dat diem toi da Vat It, tap 1 - Le Van Vinh

Chuyto dh 2

B A I T O A N G I A T R I H I ^ U D U N G

^hmmg phdp gidi ' •'' '

Six d u n g cong thuc: = U R + ( U ^ - U c )

haycos(p = — hay U = — - — hay tan(p =

Nhan xet: de'xac dinh cuang do hieu dung cua ddng dien xoay chieu, nguai ta cho

ddng dien xoay chieu va ddng dien khong ddi ciing chqy qua dien tra nao do sau thai

gian t giong nhau Cudi ciing nguai ta do dugc nhiet luang toa ra tren dien tra do

trong hai ddng dien tren la bang nhau tie do mai cd khdi niem cuang do ddng dien

hieu dung trong ddng dien xoay chieu chinh Id do Ian cua ddng dien khong dot

Bdi todn tren, thai gian dugc xet Id mot chu ky cua ddng dien xoay chieu Day la

dang todn tuy de ngan nhung hai mai la ddi vdi nhieu ban vd de thi chinh thuc

ciing chua ra thi dang nay

V l d u 2 : Chpn phat bieu sai? Mach di^n xoay chieu RLC noi tiep c6 R va C

khong doi, dang xay ra cong huong Ne'u tang L m o t l u g n g nho thi

A dien ap hieu d u n g tren tu di^n giam

B dien ap hi^u d u n g tren cuon cam giam

C cong sua't toa nhi^t tren mach giam

D dign ap higu d u n g tren di^n tra giam

^hdn tick vd hucmg ddn gidi

Ban dau mach cpng h u o n g nen Z c = Z L => Z = Z^^i^ = R

Cty TNHH MTV DWH Khang Viet

Khi tang R m o t l u g n g nho thi tong tro ciia mgch luc nay:

Z = 7 R ^ + ( Z L - Z c f t v i Z L - Z c ^ O = ^ ( Z L - Z c f > o

U ( - = I Z ( ~ = - ^ r Z ( ~ => U(- i ( Z ( - ; U = const) nen A d i i n g

R ^ + Z ^

L bien thien de Uimax k h i Z L = — Ban dau ZQ = Z L nen k h i tang L

mot lugng nho thi Z L se tang mot lugng nho v i the U L S B tang va c u tang

R ^ + Z ^

tir t u L den k h i Z L = — - — — thi ULmax v i the trong cau nay di^n ap hi^u

dung tren cugn cam tang nen dap an B sai

P = I ^ R = - ^ y y R = > P i ( R ; U = const) nen C d i i n g

U R = I R = - ^ R = > U R i ( R ; U = const) nen D d u n g

Chpn dap an B

v i d u 3: Doan mach A B g o m doan mach A N chua cuon thuan cam noi tiep

voi doan mach N B chua dien tro R va tu dien C Ggi UR, UI,, U C la dien ap hieu d u n g gii>a hai dau m o i phan tir R, L, C Biet dien ap giUa hai dau A B bien thien dieu hoa vuong pha so v o i di^n ap hai dau N B H $ thuc nao sau day dung?

CO mot cdch gidi nhanh, de hieu ma cdc ban can lini

y Id cdch sif dung vecto tritgt da dugc trinh bdy rat chi tiet trong cuon "cam nang luyen thi dai hoc

tap 1" ciia tdc gid nen cdc ban cd thetham khdo

Tif gia thict decho ta ca gian do vecto trugt:

u^g 1 U N B A A B N vuong tai B

Bi quySi dn luyftt thi dai hpc dat diem tot da V^t It, tap 1-Le Van Vinh

Vi du 4: Mpt doan mach gom di?n tra R, cupn thuan cam L va tu di^n C

mac noi tiep (trong do R, L, C la nhung gia trj hiiu han va khac 0) Dat

di^n ap xoay chieu c6 gia tri hi^u dung U, tan so i thi thay di#n ap hai

u /s

dau di§n tra R, cupn thuan cam L va ty di^n C Ian lugt UR = ^ ;

U L = y • Khi tan so dong dien la 2f thi di^n ap tren di$n tro, tren tu di^n

va tren cupc cam Ian lugt bang

A U , = M u , = ^ , U c = U ; B U , = ^ , U L = V 3 U , U C ^ ^ ;

D U R = U , U L = U , U C = U ;

'Phdn tick vd hizang dan gidi

Tu cong thuc lien he giira cac U:

Ta c6: UR = luon la hang so vi U khong doi nen khi tang tan so len

gap doi thi hieu dien the hieu dung hai dau dien tro khong thay doi

Vay chQD dap an C

Vi dlgl 5: Trong mot doan mach xoay chieu AB gom hai doan AN va NB

mac noi tiep Doan AN gom dien tro thuan R mlc noi tiep vai tu C, doan

NB chua cuon thuan cam L Khi mach dang c6 cpng huang, neu sau do

chi tang tan so cua dien ap dat vao hai dau doan mach thi ket luan nao

sau day la khong dung?

A Di^n ap hi^u dung tren doan AN tang

B Dien ap hieu dung hai dau dien tro thuan R giam

C Dong di^n trong mach cham pha hon difn ap dat vao mach hai dau

mach AB

D Cuong do hieu dung trong mach giam

Cty TNHH MTV DWH Khang Viet 'Phdn tich vd huang dan gidi

p Tong tra ciia mach: Z = coL

A Dien ap hieu dung tren doan AN: U ^ N = I - ^ A N = y y Z A N U ^ N ^

vi the dap an A sai

B Dien ap hieu dung hai dau dien tro thuan R giam la diing vi:

V I du 6: Cac doan mach xoay chieu AM, MN, NB Ian lupt chiVa cac phan

tu: bien tro R; cupn day thuan cam c6 he so tu cam L; tu dien c6 dien dung C Dat vao hai dau AB mpt di^n ap xoay chieu c6 gia trj hi^u dung

U khong doi, con so goc a' thay doi dupe De so chi cua von ke'ly tuong dat giira hai diem A, N khong phu thupc vao gia trj ciia R thi a' phai c6 gia tri

A. M =

V L C • 6 .(0 = 1

X/LC' C co = 1

,y2x:c' D co = 1

'Phdn tich vd huang dan gidi

Dion ap giua hai diem AN: V = U ^ N = I - ^ A N = — - ^ A N

Bi quyet on luy^n thi hQC dat diem tot da Vat It, tap 1 - Le Van Vitifi

Vi d u 7: Dat dien ap xoay chieu u = U \/2 coscot vao hai dau mach di^n m§c

noi tiep theo thii tu: di^n tro R, cuon thuan cam L va tu C Biet U, L, co

i<h6ng thay doi; dien dung C va dien tro R c6 the thay doi Khi C = Ci thi

dien ap hieu dung hai dau dien tro ichong phu thuoc R; khi C = C2 thi

dien ap hieu dung hai dau mach chua L va R cung khong phu thuoc R

Bieu thuc dung la

A C 2 = 0,5Ci B.C2 = Ci C C2 = 2Ci D C 2 = N / 2 C I

<Phdn tich vd hu&ng dan gidi

V l du 8: Khi mac Ian luot dipn tro thuan R cupn cam thuan L, tu di^n C vao

mot dien ap xoay chieu on dinh thi cuong dp dong dien hieu dung trong

mach Ian lupt la 2A, 1 A, 3A Khi mac noi tiep ba phan t u R,L,C do roi mSc vao

dien ap xoay chieu tren thi cuong dp dong dien hieu dung trong mach la

A 3 V2 A B 6 A C 1,20 A D 1,25 A

<Phdn tich vd huamg dan gidi

Theo bai ra, khi mac Ian lupt dien tro thuan R, cupn cam thuan L, tu di^n C vao mot di?n ap xoay chieu U khong doi nen ta c6:

Dien tro thuan: R = — = —

3

= 5 u

6 Cuong dp dong dien hieu dung luc nay: I = — = = 1,2A

-12 ( V ) vao hai dau doan mach AB thi dien

ap tuc thoi hai dau doan mach MB la u MB=120N/2 cos

Di^n ap hieu dung hai dau doan mach A M bang

Ta c6: U^B = UAM + "MB ^ "AM = "AB - "MB

V i du 10: Dat di#n ap u = U 7 2 cos(cot + <p) (V) vao hai dau mach RLC noi

tiep, cugn day thuan cam, dien d u n g C thay doi duoc K h i dien d u n g c6

C = Ci, do di?n ap hai dau cupn day, t u dien va di^n tro Ian luot

Ur, = 310V va Uc = U R = 155V K h i thay doi C = C2 de Uc2 = 155 V 2 V thi

dien ap hai dau cuon day k h i do bang

A 175,3V B 350,7V C 120,5V D 354,6V

<Phdn tick vd huang dan gidi

Theo bai ra: U L = 21)^ luon luon la hSng so

K h i C = C i ^ U = ^U^R + ( U L - f = ^155^ + (310 - 1 5 5 ) ^ = 1 5 5 V 2 V

K h i c^Ul^ - 2 U L 2 U C 2 + f^l2

= - U C 2 = - 1 5 5 7 2 = 3 5 0 , 7 V

Chpn dap an B

Nhan xet: Bai todn c6 ve de dang nhtmg neu khong cM y den he thiec U L = 2U|.;

luon luon la Mng sovi LvaR khCmg thay ddi la chung ta se c6 dap an khdc

Vl du 1 1 : Dat mot dien ap xoay chieu u = UoCoscot(V) vao hai dau mach

dien A B mac noi tiep theo t h u t y gom dien tro R, cuon day khong thuan

cam (L, r) va tu dien C voi R = r Goi N la diem nam giua di?n tro R va

cuon day, M la diem nam giCra cuon day va tu dien Dien ap tuc thoi UAM

va UNB v u o n g pha v o i nhau va c6 cung mot gia trj hi^u d u n g la 3075 V

Gia trj ciia Uo bang:

A B dien ap u = 100 N/2 coso)t(V) K h i mac ampe ke'co dien tro khong dang

ke vao hai dau doan mach M B thi ampe k e c h i ^f2/2 (A) K h i mac vao hai dau doan mach M B mot von ke dien tro rat Ion thi he so cong suat cua

mach dat gia trj cue dai So chi cua von ke'la

A lOOV B.50x/2 V C IOOV2V D 50 V

^hdn tich vd huang dan gidi

i j ! K h i mSc ampe k e v a o hai dau M B thi doan M B bj noi tat, liic nay mach chi

|| con R] noi tiep v o i L

|! T u do ta de dang tinh dupe tong tro doan A M va cam khang cua cupn day

Z I = ^ = 1 0 0 V 2 Q = ^ Z L = V Z ? - R ? = I O O Q

Khi mac von k e , h f so cong suat cue dai suy ra mach cong huong, nen ta c6

Zc = Zl = lOOQ, k h i do tong tro la Z = 2Ri= 200Q va liic nay cuong dp dong

dien se la: r = i l = M = 0 , 5 A

Z 200 Tong tro doan M B : Z ^ g = ^ R a + Z ^ = VlOO^ +100^ = 100>/2n

So chi v o n ke: V = U M B = I ' - Z M B = 0,5.100V2 = 50>y2V

^ A lOOV B I O O V 2 V C 100^/3V D 120 V

^hdn tich vd hicorng ddn gidi

Theo bai ra ta c6: cpj^ -(p,^ = ^ => - ^Pi = ^ => tancp^ tancpj = - 1

C a u 1 : Dat dien ap u =UN/2 coso)t vao hai dau doan mach R L C noi tiep, bie't

dien tro R khong doi Khi trong doan mach c6 cong huong dien thi phat

bleu nao sau day sai?

A Cuong do hieu dung cua dong dien trong mach dat gia trj Ion nhat

B Cam khang va d u n g khang cua doan mach bang nhau

C Dien ap hieu dung hai dau dien tro nho hon di?n ap hieu d u n g hai dau

doan mach

D Dien ap giua hai dau doan mach cung pha voi dien ap giua hai dau di?n tro

'Phdn tich vd hurnig ddn gidi

U , ,

-A khi mach cong huong thi Z L = Zc =^ Z = Z^i„ = R => I = I ^ ^ , = — v i the

K

Dap an A dung

B khi mach cong huong thi = ZQ v i the'dap an B diing

C khi mach cong huong thi Z L = Z ^ => Z = Z ^ j ^ = R => U R = U v i the

Dap an C sai

Z — Z

D k h i mach cong huong thi Z L = Zc ^ tancp = ^ ^ ^ = 0

(p = (p^j - (p = 0 => (pu = (Pi v i the'dap an D diing

Vay chon dap an C

C a u 2 : M o t doan mach A B gom doan A M va M B mSc noi tiep Doan A M

dien tro thuan Ri noi tiep voi tu d i ^ n c6 dien dung C i Doan mach MB

dien tro thuan R2 noi tiep tu dien c6 dien dung C2 K h i dat dien ap

u = Uocosa)t (Uo, u) khong doi) vao hai dau doan m^ch A B thi tong tro

ZAB = ZAM + ZMH He thuc lien he giOa R i , C i , R2, C2 la

A R i + R 2 = C i + C2 B R 2 C 2 = R i C i ;

i C R2Ci = RiC2 D RiR2 = CiC2 v.'-3 i t uf

'Phdn tich m huang dan gidi

Theo bai ra: Z ^ B = Z ^ M + Z ^ g ^ U A B = U ^ M + U ^ B ^ieu nay xay ra khi

U A M cung pha voi U ^ B vi the:

phai dat vao mach mot dien ap xoay chieu ccS gia trj hieu dung khong doi, c6

tan so goc w bang bao nhieu Ian coo de dien ap URL khong phu thuoc vao R ?

A 2 B.0,5 C.N/2 D l / ^ / 2

Phdn tich vd huang ddn gidi

Dien ap giua hai diem A N : U[^L = I Z R , = ^ Z R L

C a u 4 : M o t mach dien xoay chieu gom RLC noi tiep Dien tro R, cupn day

0 3 l O " ' thuan cam c6 do t u cam L = - ^ H , tu dien c6 di?n dung C = F Dien

n 6n

ap gii>a hai dau mach c6 gia trj hieu dung U khong doi va c6 tan so f thay doi Thay doi tan so f de cho dien ap hieu dung giua hai dau dien tro bang dien ap hieu dung gii>a hai dau doan mach thi f c6 gia tri la

A 5 0 V 2 H z B 100 Hz C 50 Hz D 1 0 0 ^ / 2 H z

Phdn tich vd huang ddn gidi

Theo bai ra: Thay doi tan so f de cho d i ^ n ap hieu d u n g giija hai dau di?n tro bang dien ap hieu dung gii>a hai dau doan mach khi do trong mach xay

ra cong huong dien v i the:

' 301

Cau 5: Doan mach khong phan nhanh gom mot dien tro thuan, mot cuon

cam thuan va mot tu dien dat duoi dien ap xoay chieu u = 200cos(27Tft) V

CO tan so thay doi dugc Khi tan so la fi thi dien ap hieu dung d hai dau R ij,

UR = 100 72 V Khi tan so la h thi cam khang bSng 4 Ian dung khang Ti so

fi/f2 la

A 0,25 B.0,5 C.2 D 4

^hdn tich vd hucmg ddn gidi

Theo bai ra: U = ^ = ^ = 100V2V

72 72 Khi tan so la f, thi = 10072V = U :

Khi tan so la thi:

Cau 6: Mot mach di"n xoay chieu noi tiep gom tu dien c6 dien dung C, dien tro

thuan R va cupn day c6 do tir cam L c6 dien tro thuan r Diing von keco dien

tro rat Ion Ian lugt do hai dau dien tro, hai dau cuon day va hai dau doan

mach thi so chi Ian lugt la 50V, 3072 V va 80 V Biet dien ap tuc thoi tren

cuon day som pha hon dong dien la n/4 Dien ap hieu dung tren tu bang

'Phdn tich vd hu&ng ddn gidi

Theo bai ra ta c6 gian do vecto trugt nhu hinh ve: '"^

Dien ap tuc thai tren cuon day

som pha han dong di^n la n/A

nen tarn giac NHB vuong can

U 7: Mach di^n xoay chieu R, L, C mSc noi tiep (trong do R, L khong thay

doi; C CO the thay doi) Dat vao hai dau mach dien ap u = 100cos(1007it) V

Ban dau C co gia trj sao cho dien ap hieu dung hai dau tu la Ion nhat; sau

do giam C thi cuong do dong dien hi^u dung trong mach

A giam B tang roi giam C giam roi tang D tang

^hdn tich ra huong ddn gidi , ^

C bien thien de Ucm.nv khi ZQ = ^ => ZIZQ = R^ + Z^, vt u.sj r

Tong tro cua mach liic nay: '

= ^R' Zl.zl-l(B} zl)=^Zl-R'-Zl ^

Do L, R va tan so khong doi nen khi i

C i = > Z c t ^ Z = V z ^ - R 2 - Z [ t ^ I = J ^ | Chpn dap an A

Cau 8: (Trich de thi thu Su pham Ha Npi Ian 1 nam 2013)

Dat dien ap mot chieu 121/ vao hai dau doan mach gom dien tra R mac noi

tiep voi cugn cam thuan thi cuong do dong dien trong mach la khong doi

CO gia trj 0, 24/4 Neu mac vao hai dau doan mach nay mot di^n ap xoay chieu lOOU - 50Hz thi cuong do dong dien hifu dung qua mach la l A Gia trj cua L la:

' z ^ ^ = 1 : ^ = ^ = 100

=^Zi = 7 Z 2 - R 2 =7100^-50^ = 5 0 7 3 Q ^ L = - ^ = ^ ^ = 0,275H

CO IOOTI

Cau 9: (Trich de thi thu Su pham Ha Npi Ian 1 nam 2013)

Mot doan mach gom dien tra R = 20Q mSc noi tiep voi mot cugn day Dat vao hai dau doan mach mot di^n ap xoay chieu u = 20072cosl007rt(V)

Dien ap hieu dung giiia hai dau dien tra va hai dau cugn day Ian lugt la

303

60V va 160V Di^n tro thuan va dp tu cam cua cupn day c6 gia tri tuong

ling la:

A 40Q va 0,21H B 30Q va 0,UH

C 30n va 0,28H D 40Q va 0,UH

'Phdn tich v>d hiiong ddn gidi

Yi U R + U L nen suy ra cupn day khong thuan

Theo bai ra ta c6 h? phuong trinh sau:

Cau 10: (Trich de thi thu Su pham Ha Npi lan 1 nam 2013)

Mpt bong den day toe loai llOV - 60W, c6 dp ty cam cua day toe nho khong

dang ke, mic noi tiep voi mpt cupn cam thuan c6 dp tu cam L vao nguon

difn xoay chieu c6 di^n ap hieu dung U = 120V, tan so' f = 50Hz Bong den

sang binh thuong khi dp tu cam ciia cupn cam la

A 1,11H B.0,28H C.0,89H D 0,45H

^hdn tich vd huang ddn gidi

Ta CO di^n tro cua den: R = 110' = 201,7fi

Cau 11: Mpt mach di#n xoay chieu gom cac linh ki^n ly tuong R, L, C mac

noi tiep Tan so' goc de xay ra cpng huong trong mach la coo, di^n tro R c6

the thay doi Hoi can phai dat vao mach mpt hi|u di^n the xoay chieu c6 gia

tri hi^u dyng khong doi, c6 tan so goc co bSng bao nhieu de hi^u di^n the

URC khong phy thupc vao R?

C CO = >/2coo

304

i'hdn tich v>d huang ddn gidi

t Taco: URC = I-ZRC = ^ - ^ R C

di^n dung C vao dipn ap xoay chieu u = Uocoscot thi cuong dp hipu dyng cua dong di§n qua chiing lan lupt la 4A, 6A, 2A Neu mac noi tiep cac phan tu tren

•

vao dien ap nay thi cuong dp hi?u dung ciia dong di?n qua mach la

A.12A B.4A C.6A D.2,4A

^hdn tich vd huang ddn gidi

Theo bai ra, khi mac lan lupt di|n tro thuan R, cupn cam thuan L, ty di^n C vao mpt di?n ap xoay chieu U khong doi nen ta c6:

Di^n tro thuan: R = — = —

IR 4 Cam khang cupn day: Z L = — = —

Dung khang ciia tu: Z ^ = — = —

Ic 2 Khi mac noi tiep ba phan tu R, L, C do roi mac vao difn ap xoay chieu tren thi: Tong tro cua mach liic nay:

IK;';,!

2 ^ U U^2 5

U 2 Cuong dp dong di^n hi^u dyng liic nay: I = — = = 2,4A

Z ^

12 -U

Chpn dap an D

Chnyto 3

B A I T O A N G l A T R ! T O - C THCri

PHl/CTNG PHAP

+ Cuong do dong d i ^ n trong mach: i = IQ cos(cot + (p,)

+ Hieu di?n thehai dau dien tro:

' U K =U„KCOs(cot + ( P u j J - U „ K C o s ( c o t + (p,)

+ Hieu d i ^ n thehai dau cuon cam thuan:

U L = U o L C O s ( ( o t + (p, ) = U„i_cos wt + ( P i + - =-UoLsin(cot + (Pi)

+ Hieu d i ^ n thehai d a u ty dien:

uc = U()c cos((ot + (puc ) = cosj^cot + <Pi - 2 J " Uocsin ((ot + cpi)

Bicu dien ca bon ham i ; u K ; U L ; U ( ; tren

cimg mot d u o n g tron lugng giac n h u sau

+ Cuong dp dong di?n trong mach:

i ^-1,) cos((ot + (pj) la ham cosin

cung chieu true cosin c6 chieu (+)

t u trai sang phai voi bien do la

^ ' m a x - IQ

+ Hi^u di?n the hai dau di?n tro:

" R = ^QR cos(cot + (P|) la ham cosin

=> cung chieu true cosin c6 chieu (+) t u trai sang phai voi bien do la

=>'jRtnax ='UoR thuan

+ Hi?u di^n the hai dau cupn cam: U L =-UoLsin((ot + (Pi)

La ham trir sin =:> nguoc chieu true sin nen c6 chieu (+) huong tu tren

xuong voi bien d p U L ^ J X = ^ O L va (p^^^ = 9i + ^

+ H i ^ u di^n the hai dau ty di^n: u ^ = Uocsin((ot + cpj) la ham sin => cung

chieu true sin nen co chieu (+) hudng t u d u o i len voi bien do

TTTTJ^ VIC I

m V I DU MAU

Vi du 1: Cho mach dien xoay chieu gom di?n tro thuan R, cuon day thuan

10 -3 cam L va tu dien C = F mac noi tiep Bieu thuc di?n ap gii>a hai ban

-3 = 10Q

Bieu thiic cuong do dong dien trong mach (nhanh pha hon dien ap hai dau

tu mot goc nil):

Vi du 2: Mach R noi tiep voi C Dat vao hai dau mach mot di?n ap xoay

chieu CO tan so f = 50Hz Khi dien ap tuc thai hai dau R la 20>/7 V thi

cuong do dong di?n tuc thoi la Vz A va d i ^ n ap tuc thoi hai dau tu la 45V Den khi dien ap hai dau R la 40 ^3 V thi dien ap tuc thai hai dau tu