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(D) The given wave equation is $Y = A \sin 2 \pi (n t - \frac{x}{\lambda})$.Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)

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$\frac{\pi A}{2}$

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(D) The given wave equation is $Y = A \sin 2 \pi (n t - \frac{x}{\lambda})$.Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$, we have $\omega = 2 \pi n$ and $k = \frac{2 \pi}{\lambda}$.The maximum particle velocity is $v_{p, 	ext{max}} = A \omega = A (2 \pi n) = 2 \pi n A$.The wave velocity is $v_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.According to the problem, $v_{p, 	ext{max}} = 4 v_w$.Substituting the expressions, we get $2 \pi n A = 4 (n \lambda)$.Dividing both sides by $4n$, we get $\lambda = \frac{2 \pi n A}{4 n} = \frac{\pi A}{2}$.

$A$ simple harmonic progressive wave is represented as $Y = A \sin 2 \pi (n t - \frac{x}{\lambda}) \text{ cm}$. If the maximum particle velocity is four times the wave velocity, then the wavelength of the wave is

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