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If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right) .$ Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-
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$\pi x_{0}$
$y=x_{0} \cos 2 \pi\left(\right.$ nt $\left.-\frac{x}{\lambda}\right)$
$\mathrm{y}=\mathrm{x}_{0} \cos \frac{2 \pi}{\lambda}(\mathrm{vt}-\mathrm{x}) \quad[\because \mathrm{v}=\mathrm{n} \lambda]$
$\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }=\mathrm{x}_{0} \times \frac{2 \pi}{\lambda} \mathrm{v}=2 \mathrm{v}($ given $) \therefore \lambda=\pi \mathrm{x}_{0}$
$\mathrm{y}=\mathrm{x}_{0} \cos \frac{2 \pi}{\lambda}(\mathrm{vt}-\mathrm{x}) \quad[\because \mathrm{v}=\mathrm{n} \lambda]$
$\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }=\mathrm{x}_{0} \times \frac{2 \pi}{\lambda} \mathrm{v}=2 \mathrm{v}($ given $) \therefore \lambda=\pi \mathrm{x}_{0}$
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