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Question

# solve this and explain in detail Q.1) When the maximum particle velocity for a transverse wave is 3 times the wave velocity? Take transverse wave as $y={y}_{0}\mathrm{sin}2\mathrm{\pi }\left(\mathrm{ft}-\frac{\mathrm{x}}{\mathrm{\lambda }}\right)$.

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Solution

## $DearStudent,\phantom{\rule{0ex}{0ex}}Given,\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{know}\mathrm{standard}\mathrm{equationof}\mathrm{wave}:\phantom{\rule{0ex}{0ex}}\mathrm{y}=\mathrm{r}\mathrm{sin}2\mathrm{\pi }\left(\mathrm{vt}-\frac{\mathrm{x}}{\mathrm{\lambda }}\right)\phantom{\rule{0ex}{0ex}}\mathrm{given}{\mathrm{v}}_{\mathrm{p}}=3\mathrm{v}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{o}}\mathrm{\omega }=3\left(\frac{\mathrm{\omega }}{\mathrm{k}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{o}}=\frac{3}{\mathrm{k}}\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{know},\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\phantom{\rule{0ex}{0ex}}\mathrm{thus}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{o}}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }}×3\phantom{\rule{0ex}{0ex}}ora=\lambda =\frac{2\mathrm{\pi }}{3}{y}_{o}\phantom{\rule{0ex}{0ex}}Regards$

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