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A simple harmonic progressive wave is given by $\mathrm{Y}=\mathrm{Y}_0 \sin 2 \pi$ $\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right)$. If the wave velocity is $\left(\frac{1}{8}\right)^{\text {th }}$ the maximum particle velocity, then the wavelength is
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Verified Answer
The correct answer is:
$\frac{\pi \mathrm{Y}_0}{4}$
$$
\begin{aligned}
& \text { Maximum particle velocity }=\mathrm{Y}_0 \omega=2 \pi \mathrm{n} \mathrm{Y}_0 \\
& \text { Wave velocity }=\mathrm{n} \lambda \\
& \therefore \frac{2 \pi \mathrm{n} \mathrm{Y}_0}{8}=\mathrm{n} \lambda \\
& \therefore \lambda=\frac{\pi \mathrm{Y}_0}{4}
\end{aligned}
$$
\begin{aligned}
& \text { Maximum particle velocity }=\mathrm{Y}_0 \omega=2 \pi \mathrm{n} \mathrm{Y}_0 \\
& \text { Wave velocity }=\mathrm{n} \lambda \\
& \therefore \frac{2 \pi \mathrm{n} \mathrm{Y}_0}{8}=\mathrm{n} \lambda \\
& \therefore \lambda=\frac{\pi \mathrm{Y}_0}{4}
\end{aligned}
$$
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