Question

A massless rod $PQ$ is hanging from two identical string $AP$ and $BQ$ of equal length. At point $O$ a mass of $m\text{kg}$ is hanging at length $\frac{L}{5}$ from end $P$. If standing wave is set in string $AP$ and $BQ$ and the fundamental frequency of $AP$ is equal to $n^{th}$ harmonic of $BQ$. Then the value of $n$ is

Answer 1

Let tension in $AP$ be $T_1$ and in $BQ$ be $T_2$

At equilibrium,

$T_1+T_2=mg$

Considering ${}^{\prime }{O}^{\prime }$ as reference, for rotational equilibrium,

${\tau }_{T_1}={\tau }_{T_2}$

$\Rightarrow T_1\left(\frac{L}{5}\right)=T_2\left(\frac{4L}{5}\right)$

$\Rightarrow T_1=4T_2---\left(1\right)$

In string AP, velocity of wave $V_1$ is given by,

$V_1=\sqrt{\frac{T_1}{\mu }},$

Where, $\mu :$ mass per unit length

The frequency modes in streched string is given by,

$f=\frac{nV}{2l}$

For string $AP$, given that $n=1\left(\text{fundamental frequency}\right)$,

$\Rightarrow f_1=\frac{V_1}{2l}=\frac{1}{2l}\sqrt{\frac{T_1}{\mu }}$

For string $BQ$, the velocity of wave $V_2$ is given by,

$V_2=\sqrt{\frac{T_2}{\mu }}$

For string $BQ$, $f_2=\frac{n{V}_2}{2l}$

$\Rightarrow f_2=\frac{n}{2l}\sqrt{\frac{T_2}{\mu }}$

From $\left(1\right)$, $f_2=\frac{n}{2l}\sqrt{\frac{T_1}{4\mu }}$

Given that, $f_1=f_2$

$\Rightarrow \frac{1}{2l}\sqrt{\frac{T_1}{\mu }}=\frac{n}{2l}\sqrt{\frac{T_1}{4\mu }}$

$\Rightarrow n=2$

Hence, $2$ is the correct answer.

Why this question? This question is mixed concept of rotation and wave, helps in revising concept of rotational equilibrium and standing wave.