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 length of the rope is

Answer 1

Given that the rope is in second harmonic.
We first try to find the location of a node.
Given that:
$y=0.1\sin \left(\frac{\pi x}{2}\right)\sin 12\pi t$
As we can see there is a node at $x=2m$. (because the $\sin \left(\frac{\pi x}{2}\right)$ term will go to zero).
We know that in the second harmonic there is a node at the midpoint of the string.
Hence length of string is $2\times 2=4m.$