A transverse wave is represented by y = frac{ | vedclass

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(A) The given wave equation is $y = a \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$,where amplitude $a = \frac{10}{\pi} \ cm$.The wave

Vedclass · Accepted Answer

$40$

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(A) The given wave equation is $y = a \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$,where amplitude $a = \frac{10}{\pi} \ cm$.The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{2\pi/T}{2\pi/\lambda} = \frac{\lambda}{T} = f\lambda$.The maximum particle velocity $v_{p,max}$ is given by $v_{p,max} = a\omega = a \left( \frac{2\pi}{T} \right)$.According to the problem,the wave velocity is twice the maximum particle velocity:$v = 2 v_{p,max}$Substituting the expressions:$\frac{\lambda}{T} = 2 \left( a \cdot \frac{2\pi}{T} \right)$Canceling $T$ from both sides:$\lambda = 4\pi a$Substituting the value of $a = \frac{10}{\pi} \ cm$:$\lambda = 4\pi \left( \frac{10}{\pi} \right) = 40 \ cm$.

$A$ transverse wave is represented by $y = \frac{10}{\pi} \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$. For what value of the wavelength is the wave velocity twice the maximum particle velocity (in $cm$)?

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