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

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Question

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\le

Vedclass · Accepted Answer

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

Vedclass · Answer

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 $=d(K E)=1 / 2\left(V_{y}^{2} 	imes d might)=1 / 2\left(V_{y}^{2} 	imes \mu d xight), \mu$ is the linear mass density.

$\Rightarrow 1 / 2\left(\mu 	imes \Omega^{2} 	imes A^{2} 	imes \cos ^{2}(k x-\Omega t)ight) d x$

integrating at $t=0,$ with limits as 0 and $\lambda,$ we have

$K . E=1 / 4\left(\mu 	imes \Omega^{2} 	imes A^{2} 	imes \lambdaight)$

Potential energy, $d U=1 / 2\left(\Omega^{2} 	imes y^{2} 	imes \muight) d x$

integrating at $t=0,$ with limits as 0 and $\lambda,$ we have $U=1 / 4\left(\mu 	imes \Omega^{2} 	imes A^{2} 	imes \lambdaight)$

Total energy $E=K . E+U$ $\Rightarrow E=1 / 2\left(\mu A^{2} \lambdaight)$

Therefore, for the first and fundamental frequency, energy is $E_{1}=\left(1 / 2\left(\mu A^{2} \lambdaight)ight) / n^{2}$

And clearly from the above derivation, we have, $K.E$ is half the total energy.

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

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