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 \(\left(\frac{1}{8}\right)^{\mathrm{th}}\) the maximum particle velocity then the wavelength is

Given wave equation \(\mathrm{y}=\mathrm{y}_0 \sin \pi\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right)\)
\(\therefore\) Particle velocity \(=\frac{\mathrm{dy}}{\mathrm{dt}}=\left(\mathrm{y}_0 \pi \mathrm{n}\right) \cos \pi\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right)\)
If the wave velocity is \(\left(\frac{1}{8}\right)^{\text {th }}\) of the maximum particle velocity, then,

Where \(\Lambda\) is the wavelength and \(\mathrm{f}\) is the frequency. Frequency is given by
\(f=\frac{\omega}{2 \pi}=\frac{\pi n}{2 \pi}=\frac{n}{2}\)
Therefore, by plugging \(\mathrm{f}\) into equation (1), we get the wavelength:
\(\Lambda=\frac{2 \pi y_0}{8}=\frac{\pi y_0}{4}\)