Question

A rope, under a tension of $200N$ and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by : $y=0.1\sin \left(\frac{\pi x}{2}\right)\sin 12\pi t$.
Where $x=0$ at one end of the rope, $x$ is in meters and $t$ is in seconds. The speed of the progressive waves (in $m/s$ )on the rope is

Answer 1

Given that $y=0.1\sin \left(\frac{\pi x}{2}\right)\sin 12\pi t$
Finding both $\frac{{\partial }^{2}y}{\partial t^2}$ and $\frac{{\partial }^{2}y}{\partial x^2}$
$\frac{{\partial }^{2}y}{\partial t^2}=-0.1\left(12\pi {)}^2\sin \right(\frac{\pi x}{2})\sin 12\pi t$
$\frac{{\partial }^{2}y}{\partial x^2}=-0.1\left(\frac{\pi }{2}{)}^2\sin \right(\frac{\pi x}{2})\sin 12\pi t$

Using the relation: $\frac{{\partial }^{2}y}{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^{2}y}{\partial t^2}$
$c^2=\frac{\frac{{\partial }^{2}y}{\partial t^2}}{\frac{{\partial }^{2}y}{\partial x^2}}=\frac{(12\pi {)}^2}{(\frac{\pi }{2}{)}^2}$
i.e.

$c=velocityofsound=24m{s}^{-1}$