Physics - Wave on String Question with Solution | TestHub

Given equation is $y=y_0\sin 2\pi ft$ where $A=y_0,k=\frac{2\pi }{\lambda }$ According to question, $\begin{array}{rlrl}v& =4v_p& & \\ \therefore & A\omega & =4\frac{\omega }{k}& \\ \therefore & y_0\omega & =\frac{4\omega }{k}& \\ \Rightarrow & & y_0=\frac{4}{k}& =\frac{4}{2\pi }\times \lambda \end{array}$ Therefore, wavelength $\lambda =\frac{\pi y_0}{2}$