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(D) The given wave equation is $y = A \sin(2\omega t - 2kx)$.Comparing this with the standard wave equation $y = A \sin(\omega' t - k' x)$,we have ang

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$\frac{\lambda}{4 \pi}$

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(D) The given wave equation is $y = A \sin(2\omega t - 2kx)$.Comparing this with the standard wave equation $y = A \sin(\omega' t - k' x)$,we have angular frequency $\omega' = 2\omega$ and wave number $k' = 2k$.The velocity of the particle is $v_p = \frac{dy}{dt} = A \cdot (2\omega) \cos(2\omega t - 2kx)$.The maximum particle velocity is $v_{p,max} = 2A\omega$.The wave velocity is $v_w = \frac{\omega'}{k'} = \frac{2\omega}{2k} = \frac{\omega}{k}$.Given that $v_{p,max} = v_w$,we have $2A\omega = \frac{\omega}{k}$.Thus,$A = \frac{1}{2k}$.Since the wave number $k = \frac{2\pi}{\lambda}$,we substitute this into the expression for $A$:$A = \frac{1}{2(2\pi / \lambda)} = \frac{\lambda}{4\pi}$.

$A$ transverse wave in a medium is given by $y = A \sin 2(\omega t - kx)$. It is found that the magnitude of the maximum velocity of particles in the medium is equal to the magnitude of the wave velocity. What is the value of $A$?

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