Physics - SHM Question with Solution | TestHub

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: $\Lambda =\frac{2\pi y_0}{8}=\frac{\pi y_0}{4}$