JEE Main 2022 — Waves and Sound Question with Solution

Frequency of $n^{th}$ harmonic is $\frac{nv}{2l}=400\Rightarrow \frac{nv}{0.6}=400 ...\left(1\right)$ 

And frequency of ${\left(n+1\right)}^{th}$ harmonic is $\frac{\left(n+1\right)v}{2l}=450\Rightarrow \frac{\left(n+1\right)v}{0.6}=450 ...\left(2\right)$, here, $v$ is speed of wave on string. 

From equation $\left(1\right)$, we have

$n=\frac{400\times 0.6}{v}$

Putting in equation $\left(2\right)$, 

$\Rightarrow \left[\frac{0.6\times 400}{v}+1\right]\frac{v}{0.6}=450$

$\Rightarrow \left[400+\frac{v}{0.6}\right]=450$

$\Rightarrow v=30 \mathrm{m} {\mathrm{s}}^{-1}$

Speed of wave on string is stated as $v=\sqrt{\frac{T}{\mu }}=30$, here, $\mu$ is linear mass density.

So, $\mu =\frac{T}{v^2}=\frac{2700}{900}\Rightarrow \mu =3 \mathrm{kg} {\mathrm{m}}^{-1}$