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a clamped string is oscillating in nth harmonic, then (a) total energy of oscillating will be n^2 times that of fundamental frequency. (b) total energy of oscillations will be (n-1)^2 times that of fundamental frequency. © average kinetic energy of the string over a complete oscillations is half of that of the total energy of the string (D) none of these More than one correct type.

a clamped string is oscillating in nth harmonic, then 
(a) total energy of oscillating will be n^2 times that of fundamental frequency.
(b) total energy of oscillations will be (n-1)^2 times that of fundamental frequency.
© average kinetic energy of the string over a complete oscillations is half of that of the total energy of the string 
(D) none of these 
 
More than one correct type.

Grade:11

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
2 years ago
Dear student

the total energy of oscillations will be n^2 times that of the fundamental frequency

the average kinetic energy of the string over complete oscillations is half of that of the total energy of the string.

Regards
Arun
25757 Points
2 years ago
For a sine wave, y=Asin(kxΩt)
Velocity equation for this wave is Vy=ΩAcos(kxΩt)
Kinetic energy = d(KE)=1/2(Vy2×dm)=1/2(Vy2×μdx)μ is the linear mass density.

=> 1/2(μ×Ω2×A2×cos2(kxΩt))dx
integrating at t=0, with limits as 0 and λ, we have
 
K.E=1/4(μ×Ω2×A2×λ)
Potential energy, dU=1/2(Ω2×y2×μ)dx

integrating at t=0, with limits as 0 and λ, we have
U=1/4(μ×Ω2×A2×λ)

Total energy E=K.E+U
=> E=1/2(μA2λ)
Therefore, for the first and fundamental frequency, energy is 
E1=(1/2(μA2λ))/n2
And clearly from the above derivation, we have, K.E is half the total energy.

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