A simple harmonic progressive wave is given by $\mathrm{Y}=\mathrm{Y}_0 \sin 2 \pi$ $\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda} ight)$. If the wave velocity is $\left(\frac{1}{8} ight)^{ ext {th }}$ the maximum particle velocity, then the wavelength is

Question

A simple harmonic progressive wave is given by $\mathrm{Y}=\mathrm{Y}_0 \sin 2 \pi$ $\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}ight)$. If the wave velocity is $\left(\frac{1}{8}ight)^{	ext {th }}$ the maximum particle velocity, then the wavelength is

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Maximum particle velocity }=\mathrm{Y}_0 \omega=2 \pi \mathrm{n} \mathrm{Y}_0 \ & 	ext { Wave velocity }=\mathrm{n} \lambda \ & 	herefore \frac{2 \pi \mathrm{n} \mathrm{Y}_0}{8}=\mathrm{n} \lambda \ & 	herefore \lambda=\frac{\pi \mathrm{Y}_0}{4} \end{aligned}$

A simple harmonic progressive wave is given by $\mathrm{Y}=\mathrm{Y}_0 \sin 2 \pi$…

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