Lecture physics a2 oscillations phd pham tan thi

What is an Oscillation?• Any motion that repeats itself • Described with reference to an equilibrium position where the net force is zero, and a restoring force which acts to retu

Ho Chi Minh University of Technology

What is an Oscillation?

• Any motion that repeats itself

• Described with reference to an

equilibrium position

where the net force is zero, and a

restoring force

which acts to return object to equilibrium

• Characterize by:

- Periodic (T) or frequency (f) or angular frequency (ω)

- Amplitude (A)

Simple Harmonic Oscillation/Motion

dt 2

d 2 x

x = Acos(!t + )

Total force exerts to object given by Hooke’s law

Newton’s second law of motion:

then

m

where

General solution for motion

Velocity

Acceleration

Angular frequency

It’s simplest to define our coordinate system so that the origin O is at the equilib-rium position, where the spring is neither stretched nor compressed Then x is the

x-component of the displacement of the body from equilibrium and is also the

change in the length of the spring The x-component of the force that the spring

exerts on the body is and the x-component of acceleration is given by

Figure 14.2 shows the body for three different displacements of the spring Whenever the body is displaced from its equilibrium position, the spring force tends to restore it to the equilibrium position We call a force with this character a

restoring force Oscillation can occur only when there is a restoring force

tend-ing to return the system to equilibrium

Let’s analyze how oscillation occurs in this system If we displace the body to the right to and then let go, the net force and the acceleration are to the left

(Fig 14.2a) The speed increases as the body approaches the equilibrium position O When the body is at O, the net force acting on it is zero (Fig 14.2b), but because of its motion it overshoots the equilibrium position On the other side of the

equilib-rium position the body is still moving to the left, but the net force and the accelera-tion are to the right (Fig 14.2c); hence the speed decreases until the body comes to a stop We will show later that with an ideal spring, the stopping point is at

The body then accelerates to the right, overshoots equilibrium again, and stops at the starting point ready to repeat the whole process The body is oscillating!

If there is no friction or other force to remove mechanical energy from the system, this motion repeats forever; the restoring force perpetually draws the body back toward the equilibrium position, only to have the body overshoot time after time

In different situations the force may depend on the displacement x from equi-librium in different ways But oscillation always occurs if the force is a restoring

force that tends to return the system to equilibrium

Amplitude, Period, Frequency, and Angular Frequency

Here are some terms that we’ll use in discussing periodic motions of all kinds:

The amplitude of the motion, denoted by A, is the maximum magnitude of

displacement from equilibrium—that is, the maximum value of It is always positive If the spring in Fig 14.2 is an ideal one, the total overall range of the

motion is 2A The SI unit of A is the meter A complete vibration, or cycle, is one

complete round trip—say, from A to and back to A, or from O to A, back through O to and back to O Note that motion from one side to the other

(say, to A) is a half-cycle, not a whole cycle.

The period, T, is the time for one cycle It is always positive The SI unit is the

second, but it is sometimes expressed as “seconds per cycle.”

The frequency, is the number of cycles in a unit of time It is always posi-tive The SI unit of frequency is the hertz:

This unit is named in honor of the German physicist Heinrich Hertz (1857–1894), a pioneer in investigating electromagnetic waves

The angular frequency, is times the frequency:

We’ll learn shortly why is a useful quantity It represents the rate of change of

an angular quantity (not necessarily related to a rotational motion) that is always measured in radians, so its units are Since is in we may regard the number as having units

From the definitions of period T and frequency we see that each is the

recip-rocal of the other:

(14.1)

f = T1 T = 1ƒ (relationships between frequency and period)

ƒ

rad>cycle

2p

cycle>s, ƒ

rad>s

v

v = 2pƒ

2p v,

1 hertz = 1 Hz = 1 cycle>s = 1 s-1 ƒ,

-A -A,

-A

ƒxƒ

x = A,

x = -A.

x = A

a x = F x >m F x a x

,

x , 0: glider displaced

to the left from the equilibrium position.

F x.0, so a x.0:

compressed spring pushes glider toward equilibrium position.

F x

a x

F x

x 5 0: The relaxed spring exerts no force on the

glider, so the glider has zero acceleration.

(b)

y

x n

mg y

(a)

x x

y

x n

mg

y

x 0: glider displaced

to the right from the equilibrium position.

F x,0, so a x,0:

stretched spring pulls glider toward equilibrium position.

F x

a x

F x

(c)

x x

y

x n

mg y

14.2 Model for periodic motion When the body is displaced from its equilibrium position at the spring exerts a restoring force back toward the equilib-rium position.

x = 0,

Application Wing Frequencies

The ruby-throated hummingbird (Archilochus colubris) normally flaps its wings at about

50 Hz, producing the characteristic sound that gives hummingbirds their name Insects can flap their wings at even faster rates, from

330 Hz for a house fly and 600 Hz for a mos-quito to an amazing 1040 Hz for the tiny biting midge.

Phase of the motion

Period and Frequency

1 f

or

Simple Harmonic Motion

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

The relation between

14.2 Simple Harmonic Motion 445

To find we divide Eq (14.17) by Eq (14.14) This eliminates A and gives an

equation that we can solve for

(14.18)

It is also easy to find the amplitude A if we are given and We’ll sketch

the derivation, and you can fill in the details Square Eq (14.14); then divide

Eq (14.17) by square it, and add to the square of Eq (14.14) The right side

will be which is equal to The final result is

(14.19)

Note that when the body has both an initial displacement and a nonzero initial

velocity the amplitude A is not equal to the initial displacement That’s

rea-sonable; if you start the body at a positive but give it a positive velocity it

will go farther than x0 before it turns and comes back.

v 0 x,

x0

v 0 x,

x0

A =

Bx02 +

v 0 x2

v2 (amplitude in SHM)

A2.

A21sin 2 f + cos 2 f 2,

v,

v 0 x.

x0

f = arctana - v 0 x

vx0b (phase angle in SHM)

v 0 x

x0 = -vAsinf Acos f = -vtanf

f:

f,

x x x x x x x x x

a x! 2amax

a x5 2amax

a x5amax

x 5 2A x 50 x 5A

v x5 0

v x5 0

v x5 0

a x

v x

a x

v x

a x

v x

a x

v x

2A 2A/2 0 A/2 A

a x5 0

a x5 0

x

v x5 2vmax

v x5vmax

14.13 How x-velocity and

x-acceleration vary during one cycle

of SHM. a x

v x

Problem-Solving Strategy 14.1 Simple Harmonic Motion I: Describing Motion

IDENTIFY the relevant concepts:An oscillating system undergoes

simple harmonic motion (SHM) only if the restoring force is

directly proportional to the displacement.

SET UP the problemusing the following steps:

1 Identify the known and unknown quantities, and determine

which are the target variables.

2 Distinguish between two kinds of quantities Properties of the

system include the mass m, the force constant k, and quantities

derived from m and k, such as the period T, frequency and

angular frequency These are independent of properties of the

motion, which describe how the system behaves when it is set

into motion in a particular way; they include the amplitude A,

maximum velocity and phase angle , and values of x,

and at particular times.

3 If necessary, define an x-axis as in Fig 14.13, with the

equilib-rium position at x = 0.

a x

v x, f

vmax,

v.

ƒ,

EXECUTEthe solutionas follows:

1 Use the equations given in Sections 14.1 and 14.2 to solve for the target variables.

2 To find the values of x, and at particular times, use Eqs.

(14.13), (14.15), and (14.16), respectively If the initial position and initial velocity are both given, determine and A

from Eqs (14.18) and (14.19) If the body has an initial posi-tive displacement but zero initial velocity then the amplitude is and the phase angle is If it has

an initial positive velocity but no initial displacement

the amplitude is and the phase angle is

Express all phase angles in radians.

EVALUATEyour answer:Make sure that your results are consistent.

For example, suppose you used and to find general

expres-sions for x and at time t If you substitute into these expres-sions, you should get back the given values of x0and v 0x.

t = 0

v x

v 0x

x0

f = -p>2.

A = v 0x>v

1x0 = 02,

v 0x

f = 0.

A = x0

1v 0x = 02,

x0

f

v 0x

x0

a x

v x,

Example 14.3 Describing SHM

We give the glider of Example 14.2 an initial displacement x0 = SOLUTION

IDENTIFY and SET UP: As in Example 14.2, the oscillations are

SHM We use equations from this section and the given values k !

200 N m, m ! 0.50 kg, and to calculate the target variables A and and to obtain expressions for x, and

Continued

a x

v x, f

v 0x

x0,

>

and an initial velocity ! (a) Find the

period, amplitude, and phase angle of the resulting motion (b) Write

equations for the displacement, velocity, and acceleration as

func-tions of time.

+0.40 m>s.

v 0x

+0.015 m

Angular Frequency

The

We may also define an angular frequency ω , in radians per second, to

The position of an object oscillating with simple harmonic motion can then

!(in rad/s) = 2⇡

T = 2⇡f x(t) = Acos!t

change f

Displacement as function of time in SHM

Changing m, A or k changes the graph of x versus t:

m

Mechanical Energy in Simple Harmonic Motion

Potential Energy, U

Consider the oscillation of a spring as a SHM, the potential energy

of the spring is given by

2 kx

2 where k = m! 2

U = 1

2 m!

The potential energy in terms of time, t, is given by

2 x 2 where x = Asin(!t + )

Mechanical Energy in Simple Harmonic Motion

Kinetic Energy, K

The kinetic energy of an object in SHM is given by

where

The kinetic energy in terms of time, t, is given by

where

K = 1

2 mv

2

K = 1

2 m!

2 (A 2 x 2 )

2 m!

2 A 2 cos 2 (!t + )

Mechanical Energy in Simple Harmonic Motion

Total Energy, E

The total energy of a body in SHM is the sum of its kinetic energy, K

and its potential energy, U

The equation of total energy in SHM is given by

E = K + U

From the principle of conservation of energy, this total energy is

always constant in a closed system

E = K + U = constant

2 m!

2 m!

2

OR

Mechanical Energy in Simple Harmonic Motion

E = 1

2 mv

2 + 1

2 kx

2 = 1

2 kA

2 = 1

2 mv

2 max

Conservation of Energy

14.3 Energy in Simple Harmonic Motion 447

(Recall that Hence our expressions for displacement and

velocity in SHM are consistent with energy conservation, as they must be.

We can use Eq (14.21) to solve for the velocity of the body at a given

dis-placement x:

(14.22)

The sign means that at a given value of x the body can be moving in either

direction For example, when

Equation (14.22) also shows that the maximum speed occurs at

Using Eq (14.10), we find that

(14.23)

This agrees with Eq (14.15): oscillates between and

Figure 14.14 shows the energy quantities E, K, and U at and

Figure 14.15 is a graphical display of Eq (14.21); energy (kinetic,

potential, and total) is plotted vertically and the coordinate x is plotted horizontally.

x = ! A.

x = ! A>2,

x = 0,

+vA.

-vA

vx

vmax = A m k A = vA

v = 2k>m,

x = 0.

vmax

vx = ! A m k

B A2 - a!

A

2 b

2

= ! A 3 4 A m k A

x = ! A>2,

!

vx = !

A

k

m 2 A

2 - x2

vx

sin2a + cos2a = 1.)

E is all potential energy.

E is all potential energy.

E is partly potential, partly kinetic energy.

E is partly potential, partly kinetic energy.

E is all kinetic energy.

ax 5 amax ax 5 2 amax

vx 5 6 vmax

ax 5 amax ax 5 0

vmax

1 2

Å

x

3

1

E 5 K 1 U E 5 K 1 U E 5 K 1 U E 5 K 1 U E 5 K 1 U

equally spaced positions are not equally spaced in time.

The total mechanical energy E is constant.

(a) The potential energy U and total mechanical

energy E for a body in SHM as a function of

displacement x

Energy

x

U 5 kx12 2

E K

U

At x 5 6A the energy is all potential; the kinetic

energy is zero.

At x 5 0 the energy is all kinetic;

the potential energy is zero.

At these points the energy is half kinetic and half potential.

(b) The same graph as in (a), showing

kinetic energy K as well

Energy

x

E 5 K 1 U

2 A

U K

energy U, and total mechanical energy E

as functions of position for SHM At each

value of x the sum of the values of K and

U equals the constant value of E Can you

show that the energy is half kinetic and

half potential at x = ! 212A ?