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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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Verified Answer
The correct answer is:
$\lambda=\frac{\pi x_0}{2}$
Hints: Maximum particle velocity $\left(\mathrm{V}_{\mathrm{p}}\right)=\mathrm{A} \omega=2 \pi n x_0$
Wave velocity $\left(\mathrm{V}_\omega\right)=n \lambda$
Here, $\mathrm{V}_{\mathrm{P}}=4 \mathrm{~V}_\omega$
$$
2 \pi n x_0=4 n \lambda
$$
$$
\lambda=\frac{\pi}{2} x_0
$$
Wave velocity $\left(\mathrm{V}_\omega\right)=n \lambda$
Here, $\mathrm{V}_{\mathrm{P}}=4 \mathrm{~V}_\omega$
$$
2 \pi n x_0=4 n \lambda
$$
$$
\lambda=\frac{\pi}{2} x_0
$$
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