When a string is excited by bowing, plucking, or striking, the resulting vibration can be considered to be a combination of several modes of vi- bration. For example, if the string is plucked at its center, the resulting vibration will consist of the fundamental plus the odd-numbered harmon- ics. Fig. 2.5 illustrates how the modes associated with the odd-numbered harmonics, when each is present in the right proportion, add up at one instant in time to give the initial shape of the center-plucked string. Modes 3, 7, 11, etc., must be opposite in phase from modes, 1, 5, and 9 in order to give maximum displacement at the center, as shown at the top. Finding the normal mode spectrum of a string, given its initial displacement, calls for frequency analysis or Fourier analysis.
2.8. Plucked String: Time and Frequency Analyses 41
~"~""'' amplitude Phase Relative
~1 1 + "0 ., Spectrum
2 0 a
~ 3 1 :a e
9
4 0 oS 00
~ 5 .L 2S + ....1 0
6 0 f.
~ 7 .L 49 Frequency
FIGURE 2.5. Frequency analysis of a string plucked at its center. Odd-numbered modes of vibration add up in appropriate amplitude and phase to give the shape of the string.
Since all the modes shown in Fig. 2.5 have different frequencies of vi- bration, they quickly get out of phase, and the shape of the string changes rapidly after plucking. The shape of the string at each moment can be ob- tained by adding the normal modes at that particular time, but it is more difficult to do so because each of the modes will be at a different point in its cycle. The resolution of the string motion into two pulses that propagate in opposite directions on the string, which we might call time analysis, is illustrated in Fig. 2.6. If the string is plucked at a point other than its cen- ter, the spectrum or recipe of the constituent modes is different, of course.
For example, if the string is plucked i of the distance from one end, the spectrum of mode amplitudes shown in Fig. 2.7 is obtained. Note that the 5th harmonic is missing. Plucking the string i of the distance from the end suppresses the 4th harmonic, etc. (In Fig. 2.5, plucking it at ~ the distance eliminated the 2nd harmonic as well as other even-numbered ones.)
A time analysis of the string plucked at i of its length is shown in Fig. 2.8. A bend racing back and forth within a parallelogram boundary can be viewed as the resultant of two pulses (dashed lines) traveling in opposite directions. Each of these pulses can be described by one term in d'Alembert's solution [Eq. (2.5)].
Each of the normal modes described in Eq. (2.13) has two coefficients An and Bn whose values depend upon the initial excitation of the string.
These coefficients can be determined by Fourier analysis. Multiplying each side of Eq. (2.14) and its time derivative by sin m1rxj L and integrating from 0 to L gives the following formulae for the Fourier coefficients:
A 2 {L . ( 0) . n7rX d
n = WnL Jo y x, sm L x, (2.17)
2 {L . n1rx
Bn = L Jo y(x, 0) sm Ldx. (2.18)
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42 2. Continuous Systems in One Dimension: Strings and Bars
-+-::..c::--.
...,.,...- --
...,.,...-- ---
-- --
---::;;;:-
<..-:E '
--- ---
---- --
~--... --- -- --- -- --- ------~ ---~
--... ' -----::::-c.------- --. /
... :;::~!!':--== ---t:-... "
----::!-'---- --~--~ --- ---
t=O
t = !T
I =;\T
I= foT
t=jT
t = tT
FIGURE 2.6. Time analysis of the motion of a string plucked at its midpoint through one half cycle. Motion can be thought of as due to two pulses traveling in opposite directions.
r- I'
'\ I\
I ....
I I'
v \ \ y
0 f 5( !Of 15f
FIGURE 2. 7. Spectrum of a string plucked one-fifth of the distance from one end.
O.OST
O.JT
0.4T
/ ~- O.ST
2.8. Plucked String: Time and Frequency Analyses 43
---
~
-- _ ..
- --./ /
--
__ /
~
+-
---
F(t) 1
I I I I ~
0 0.2 0.4 0.6 0.8 .1.0 t/T
FIGURE 2.8. Time analysis through one half cycle of the motion of a string plucked one-fifth of the distance from one end. The motion can be thought of as due to two pulses [representing the two terms in Eq. (2.5)] moving in oppo- site directions (dashed curves). The resultant motion consists of two bends, one moving clockwise and the other counterclockwise around a parallelogram. The normal force on the end support, as a function of time, is shown at the bottom.
Using these formulae, we can calculate the Fourier coefficients for the string of length L plucked with amplitude h at one-fifth of its length, as shown in the time analysis in Fig. 2.8. The initial conditions are
iJ(x, 0) = 0,
y(x, 0) = -y;x, 5h 0 ~ x ~ L/5, (2.19)
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44 2. Continuous Systems in One Dimension: Strings and Bars
= 5: (1- z).
Using the first condition in Eq. (2.17) gives An = 0. Using the second condition in Eq. (2.18) gives
_ ~ {L/5 5h , n7l'X ~ {L 5h
Bn - L Jo L X sm L dx + L J L/5 4 ( 1- L X) smLdx , n1l'X
25h . n7l'
= 2n271'2 sm 5' (2.20)
The individual Bn's become: B1 = 0.7444h, B2 = 0.3011h, Ba = 0.1338h,
B4 = 0.0465h, B5 = 0, Ba = -0.0207h, etc. Figure 2.7 shows 20 log IBnl for n = 0 to 15. Note that Bn = 0 for n = 5, 10, 15, etc., which is the signature of a string plucked at 1/5 of its length.