Physics - Wave on String Question with Solution | TestHub

Given that, the displacement of a particle is $$y=A\sin (\omega t-kx)...(i)$$ The particle velocity $$v_p=\frac{dy}{dt}...(ii)$$ Now, on differentiating Eq. (i) w.r.t, $t$, $$\begin{array}{rl}& \frac{dy}{dt}=A\cos (\omega t-kx)\cdot \omega \\ \Rightarrow & \frac{dy}{dt}=A\omega \cos (\omega t-kx)\end{array}$$ From Eq. (ii) $$\Rightarrow v_p=A\omega \cos (\omega t-kx)$$ For maximum particle velocity, $$\cos (\omega t-kx)=1$$ So, $v_p=A\omega \times 1$ $\Rightarrow$ $$v_p=A\omega$$