Physics - Wave on String Question with Solution | TestHub

The given wave equation is $y(x,t)=A\sin (kx-\omega t)$. Comparing with $y(x,t)=0.05\sin (\pi x-4\pi t)$, we have amplitude $A=0.05\text{ m}$ and angular frequency $\omega =4\pi \text{ rad/s}$. The particle velocity $v_p=\frac{\partial y}{\partial t}$.$$v_p=\frac{\partial }{\partial t}[0.05\sin (\pi x-4\pi t)]=0.05(-\cos (\pi x-4\pi t))(-4\pi )$$ $$v_p=0.2\pi \cos (\pi x-4\pi t)$$ The maximum particle velocity is $v_{p,max}=A\omega$.$$v_{p,max}=(0.05\text{ m})(4\pi \text{ rad/s})=0.2\pi \text{ m/s}$$