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A clamped string is oscillating in $n^{th}$ harmonic, then
total energy of oscillations will be $n^2$ times that of fundamental frequency
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.
both $(A)$ and $(C)$
Solution
For a sine wave, $y=A \sin (k x-\Omega t)$
Velocity equation for this wave is $V_{y}=\Omega A \cos (k x-\Omega t)$ Kinetic energy $=d(K E)=1 / 2\left(V_{y}^{2} \times d m\right)=1 / 2\left(V_{y}^{2} \times \mu d x\right), \mu$ is the linear mass density.
$\Rightarrow 1 / 2\left(\mu \times \Omega^{2} \times A^{2} \times \cos ^{2}(k x-\Omega t)\right) d x$
integrating at $t=0,$ with limits as 0 and $\lambda,$ we have
$K . E=1 / 4\left(\mu \times \Omega^{2} \times A^{2} \times \lambda\right)$
Potential energy, $d U=1 / 2\left(\Omega^{2} \times y^{2} \times \mu\right) d x$
integrating at $t=0,$ with limits as 0 and $\lambda,$ we have $U=1 / 4\left(\mu \times \Omega^{2} \times A^{2} \times \lambda\right)$
Total energy $E=K . E+U$ $\Rightarrow E=1 / 2\left(\mu A^{2} \lambda\right)$
Therefore, for the first and fundamental frequency, energy is $E_{1}=\left(1 / 2\left(\mu A^{2} \lambda\right)\right) / n^{2}$
And clearly from the above derivation, we have, $K.E$ is half the total energy.
