The equation of vibration of a stretched stri | vedclass

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(C) For the $5^{	ext{th}}$ harmonic, a string of length $L$ fixed at both ends vibrates in $5$ loops.The relationship between length $L$ and waveleng

Vedclass · Accepted Answer

$12.5$

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(C) For the $5^{	ext{th}}$ harmonic, a string of length $L$ fixed at both ends vibrates in $5$ loops.The relationship between length $L$ and wavelength $\lambda$ is given by $L = \frac{5\lambda}{2}$.Comparing the given equation $Y = 3 \sin(0.4x) \cos(200\pi t)$ with the standard wave equation $Y = A \sin(kx) \cos(\omega t)$, we find the wave number $k = 0.4 	ext{ rad/cm}$.Since $k = \frac{2\pi}{\lambda}$, we have $\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.4} = 5\pi 	ext{ cm}$.Substituting the value of $\lambda$ into the length equation: $L = \frac{5 	imes 5\pi}{2} = \frac{25\pi}{2} = 12.5\pi 	ext{ cm}$.

The equation of vibration of a stretched string fixed at both ends and vibrating in $5^{\text{th}}$ harmonic is $Y = 3 \sin(0.4x) \cos(200\pi t)$, where '$x$' and '$Y$' are in $cm$ and '$t$' is in seconds. The length of the string is: (in $\pi \text{ cm}$)

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