11
Vibrations Chapter and Waves
77. (a) The overtones are given by f n
nf1 , n
2, 3, 4
f3 3 392 Hz 1180 Hz
G : f2
2 392 Hz
784 Hz
A : f 2 2 440 Hz 880 Hz f3 3 440 Hz 1320 Hz (b) If the two strings have the same length, they have the same wavelength. The frequency difference is then due to a difference in wave speed caused by different masses for the strings. FT 2 2 mG L fG vG vG mG mA fA 440 1. 26 f A vA vA mG mA fG 392 FT
mA L (c) If the two strings have the same mass per unit length and the same tension, then the wave speed on both strings is the same. The frequency difference is then due to a difference in wavelength. For the fundamental, the wavelength is twice the length of the string. fG v G LG 2 LA f A 440 A 1 . 12 fA v A 2 LG LA f G 392 G (d) If the two strings have the same length, they have the same wavelength. The frequency difference is then due to a difference in wave speed caused by different tensions for the strings.
FTG fG fA vG vA vG vA mL FTA mL
78. (a) Since the cord is not accelerating to the left or right, the tension in the cord must be the same everywhere. Thus the tension is the same in the two parts of cord. the The speed difference will then be due to the different mass densities of the two parts of the cord. Let the symbol represent the mass per unit length of each part of the cord.
vH vL FT FT
vf vf
H L H
FTG FTA
FTG FTA A
fG fA
2
392 440
2
0. 79 4
L
(b) The wavelength ratio is found as follows.
H L H L
vH vL
L H
The two frequencies must be the same for the cord to remain continuous at the boundary. If the two parts of the cord oscillate at different frequencies, the cord cannot stay in one piece, because the two parts would be out of phase with each other at various times. (c) Since H , we see that H , and so the wavelength is greater in the lighter cord . L L 79. (a) The maximum speed is given by vmax 2 f A 2 264 Hz 1.8 10 3 m (b) The maximum acceleration is given by
amax 4
2
3.0 m s .
f 2A 4
2
264 Hz
2
1.8 10 3 m
5.0 103 m s2 .
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