Wave Motion 2 Question 41

41. A string of mass per unit length μ is clamped at both ends such that one end of the string is at x=0 and the other is at x=l. When string vibrates in fundamental mode amplitude of the mid-point O of the string is a, and tension in the string is T. Find the total oscillation energy stored in the string.

(2003, 4M)

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Answer:

Correct Answer: 41. π2a2T4l

Solution:

  1. l=λ2 or λ=2l,k=2πλ=πl

The amplitude at a distance x from

x=0 is given by A=asinkx

Total mechanical energy at x of length dx is

dE=12(dm)A2ω2=12(μdx)(asinkx)2(2πf)2

or

dE=2π2μf2a2sin2kxdx

Here,

f2=v2λ2=(Tμ)(4l2) and k=πl

Substituting these values in Eq. (i) and integrating it from x=0 to x=l, we get total energy of string

E=π2a2T4l



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