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(A) The given wave equation is $y = A \sin^2 (\omega t - kx)$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,we can re

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$\lambda / 2\pi$

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(A) The given wave equation is $y = A \sin^2 (\omega t - kx)$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,we can rewrite the equation as:$y = \frac{A}{2} [1 - \cos(2\omega t - 2kx)]$.The particle velocity $v_p$ is given by the derivative of $y$ with respect to time $t$:$v_p = \frac{\partial y}{\partial t} = \frac{A}{2} [2\omega \sin(2\omega t - 2kx)] = A\omega \sin(2\omega t - 2kx)$.The maximum particle velocity is $(v_p)_{\max} = A\omega$.The wave velocity $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$ in the argument of the function:$v = \frac{2\omega}{2k} = \frac{\omega}{k}$.Since $k = \frac{2\pi}{\lambda}$,we have $v = \frac{\omega}{2\pi / \lambda} = \frac{\omega \lambda}{2\pi}$.According to the problem,$(v_p)_{\max} = v$:$A\omega = \frac{\omega \lambda}{2\pi}$.Solving for $A$,we get $A = \frac{\lambda}{2\pi}$.

$A$ transverse wave in a medium is described by the equation $y = A \sin^2 (\omega t - kx)$. The magnitude of the maximum velocity of particles in the medium will be equal to that of the wave velocity,if the value of $A$ is ($\lambda$ = wavelength of wave).

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