Q44 / 117 · NTA Abhyas (JEE Main) 2020 · Physics · Waves and Sound

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 the string vibrates in fundamental mode, the 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.

  • Correct answer
    A
    π 2 a 2 T 4 L
  • B
    π a T 2 L
  • C
    - π 2 a 2 T 3 L
  • D
    π 2 a T 6 L

Solution

λ2=L

λ=2L

τ=μV2

The amplitude at a distance x from the origin is given by
A = asinkx
consider an element of mass dm and length dx of string at a distance x from the origin
The total energy of this element = its maximum KE

= 1 2 d m ω 2 A 2

= 1 2 μ d x⁡ 4 π 2 f 2 a 2 sin 2 k⁡ x

(Because μ= dm dx and  ω=2πf  and it is given that   A = a sinkx   )

total energy of string = ∫ 0 L 2 π 2 μ f 2 a 2 sin 2 k⁡ x d x

= π 2 μ f 2 a 2 [ x - sin 2 k⁡ x 2 ] 0 L

= π 2 μ f 2 a 2 L

= π 2 μ V 2 λ 2 a 2 L = π 2 4 L 2 T a 2 L = π 2 a 2 T 4 L

Hence, the correct answer is option: 'A'.

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