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(D) The given wave equation is $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$.Comparing this with the standard wave equation $y = a \sin \frac{2\pi}{\la

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$\lambda = \pi y_0$

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(D) The given wave equation is $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$.Comparing this with the standard wave equation $y = a \sin \frac{2\pi}{\lambda} (vt - x)$,we identify the amplitude $a = y_0$ and wave velocity $v_{wave} = v$.The particle velocity $v_p$ is given by the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 \cdot \frac{2\pi v}{\lambda} \cos \frac{2\pi}{\lambda} (vt - x)$.The maximum particle velocity is $(v_{max})_{particle} = y_0 \cdot \frac{2\pi v}{\lambda}$.According to the problem,$(v_{max})_{particle} = 2 \cdot v_{wave}$.Substituting the values: $\frac{y_0 \cdot 2\pi v}{\lambda} = 2v$.Canceling $v$ from both sides: $\frac{2\pi y_0}{\lambda} = 2$.Solving for $\lambda$: $\lambda = \pi y_0$.

$A$ transverse wave is represented by the equation $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$. For what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity?

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