A clamped string is oscillating in the n^{th} | vedclass

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(D) For a standing wave,the total energy $E$ is proportional to the square of the frequency $\Omega^2$ and the square of the amplitude $A^2$. Since $\

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both $(A)$ and $(C)$

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(D) For a standing wave,the total energy $E$ is proportional to the square of the frequency $\Omega^2$ and the square of the amplitude $A^2$. Since $\Omega_n = n \Omega_1$,the energy $E_n \propto n^2 E_1$. Thus,the total energy of the $n^{th}$ harmonic is $n^2$ times the energy of the fundamental frequency.For any harmonic oscillator (including segments of a string),the average kinetic energy over a complete cycle is equal to the average potential energy. Since the total energy $E = K.E._{avg} + P.E._{avg}$ and $K.E._{avg} = P.E._{avg}$,it follows that $K.E._{avg} = E / 2$.Therefore,both statements $(A)$ and $(C)$ are correct.

$A$ clamped string is oscillating in the $n^{th}$ harmonic,then

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