A transverse wave is represented by y = 2 sin | vedclass

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Question

(A) The given wave equation is $y = A \sin(\omega t - kx)$,where amplitude $A = 2 \ cm$.The maximum particle velocity is given by $v_{p,max} = A\omega

Vedclass · Accepted Answer

$4 \pi$

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(A) The given wave equation is $y = A \sin(\omega t - kx)$,where amplitude $A = 2 \ cm$.The maximum particle velocity is given by $v_{p,max} = A\omega$.The wave velocity (phase velocity) is given by $v_w = \frac{\omega}{k}$.According to the problem,the wave velocity is equal to the maximum particle velocity:$v_w = v_{p,max}$$\frac{\omega}{k} = A\omega$Canceling $\omega$ from both sides,we get:$\frac{1}{k} = A$Since $k = \frac{2\pi}{\lambda}$,we substitute this into the equation:$\frac{\lambda}{2\pi} = A$$\lambda = 2\pi A$Given $A = 2 \ cm$,we have:$\lambda = 2\pi(2) = 4\pi \ cm$.

$A$ transverse wave is represented by $y = 2 \sin(\omega t - kx) \ cm$. The value of wavelength (in $cm$) for which the wave velocity becomes equal to the maximum particle velocity is:

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