A string tied between x=0 and x=2.45m vibrates in fundamental mode. The amplitude of vibration is 1m, tension in the string is 9.8N and mass per unit length of the string is 1g. Then, total energy of the string is
A
π2J
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B
2π2J
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C
π22J
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D
πJ
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Solution
The correct option is Aπ2J 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=asinkx Mechnical energy at x of elemental length dx is dE=12(dm)A2ω2=12(μdx)asinkx)2(2πv)2 ⇒dE=2π2μf2a2sin2kxdx 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=9.8N,a=1m and l=2.45m We have , E=π2Ta24l=π2×9.8×124×2.45=π2J