The vibrations of a string of length 60 , cm | vedclass

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(B) The standard equation for a stationary wave in a string fixed at both ends is $y = 2A \sin(kx) \cos(\omega t)$.Comparing the given equation $y = 2

Vedclass · Accepted Answer

$16$

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(B) The standard equation for a stationary wave in a string fixed at both ends is $y = 2A \sin(kx) \cos(\omega t)$.Comparing the given equation $y = 2 \sin \left( \frac{4 \pi x}{15} ight) \cos (96 \pi t)$ with the standard form,we identify the wave number $k$ as $k = \frac{4 \pi}{15} \, cm^{-1}$.The wave number is related to the wavelength $\lambda$ by $k = \frac{2 \pi}{\lambda}$.Thus,$\frac{2 \pi}{\lambda} = \frac{4 \pi}{15}$,which gives $\lambda = \frac{15}{2} = 7.5 \, cm$.For a string of length $L$ fixed at both ends,the length is related to the number of loops $p$ and the wavelength $\lambda$ by $L = p \frac{\lambda}{2}$.Given $L = 60 \, cm$ and $\lambda = 7.5 \, cm$,we have $60 = p 	imes \frac{7.5}{2}$.$60 = p 	imes 3.75$.$p = \frac{60}{3.75} = 16$.Therefore,the number of loops formed is $16$.

The vibrations of a string of length $60 \, cm$ fixed at both ends are represented by the equation $y = 2 \sin \left( \frac{4 \pi x}{15} \right) \cos (96 \pi t)$,where $x$ and $y$ are in $cm$. The maximum number of loops that can be formed in it is

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