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Question

A string tied between x=0 and x=2.45 m vibrates in fundamental mode. The amplitude of vibration is 1 m, tension in the string is 9.8 N and mass per unit length of the string is 1 g. Then, total energy of the string is


A
π2 J
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B
2π2 J
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C
π22 J
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D
π J
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Solution

The correct option is A π2 J
Here,
l=λ2 or λ=2l
Angular wave number k=2πλ=2π2l=πl
The amplitude of vibration at a distance x from x=0 is given by A=asin kx
Mechnical energy at x of elemental length dx is
dE=12(dm)A2ω2=12(μdx)asin kx)2(2πv)2
dE=2π2μf2a2sin2kx dx
Using , v=fλ and the data given,

dE=2π2T4l2a2sin2{(πxl)}dx
Total energy of the string.
E=dE=l02π2T4l2a2sin2(πxl)dx=π2Ta24l

Substituting the values of T=9.8 N,a=1 m and l=2.45 m
We have , E=π2Ta24l=π2×9.8×124×2.45=π2 J

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