A longitudinal wave is represented by x=x_0 s | vedclass

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(C) The given wave equation is $x = x_0 \sin 2 \pi (nt - x/\lambda)$.Comparing this with the standard form $x = A \sin (\omega t - kx)$, we have ampli

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$\lambda=\frac{\pi x_0}{2}$

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(C) The given wave equation is $x = x_0 \sin 2 \pi (nt - x/\lambda)$.Comparing this with the standard form $x = A \sin (\omega t - kx)$, we have amplitude $A = x_0$, angular frequency $\omega = 2 \pi n$, and wave number $k = 2 \pi / \lambda$.The maximum particle velocity $V_p$ is given by $V_p = A \omega = x_0 (2 \pi n) = 2 \pi n x_0$.The wave velocity $V_w$ is given by $V_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.According to the problem, $V_p = 4 V_w$.Substituting the expressions, we get $2 \pi n x_0 = 4 (n \lambda)$.Dividing both sides by $n$, we get $2 \pi x_0 = 4 \lambda$.Therefore, $\lambda = \frac{2 \pi x_0}{4} = \frac{\pi x_0}{2}$.

$A$ longitudinal wave is represented by $x=x_0 \sin 2 \pi(n t-x / \lambda)$. The maximum particle velocity will be four times the wave velocity if:

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