JEE Main 2023 — Waves and Sound Question with Solution

The formula to calculate the fundamental frequency in a stretched string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }} ...\left(1\right)$

For the first mass, it can be written that

$f_1=\frac{1}{2L}\sqrt{\frac{m_{1}g}{\mu }} ...\left(2\right)$

And, for the second mass, it can be written that

$f_2=\frac{1}{2L}\sqrt{\frac{m_{2}g}{\mu }} ...\left(3\right)$

Divide equation (3) by equation (2) and simplify to obtain the required mass.

$\begin{array}{rcl}\frac{f_2}{f_1}& =& \frac{\frac{1}{2L}\sqrt{\frac{m_{2}g}{\mu }}}{\frac{1}{2L}\sqrt{\frac{m_{1}g}{\mu }}}\\ & =& \sqrt{\frac{m_2}{m_1}}\\ & \Rightarrow & \frac{{\displaystyle m_1}}{{\displaystyle m_2}}={\left(\frac{f_1}{f_2}\right)}^2\\ & \Rightarrow & m_2=m_1{\left(\frac{{\displaystyle f_2}}{{\displaystyle f_1}}\right)}^2 ...\left(4\right)\end{array}$

Substitute the values of the known parameters into equation (4) to calculate the required mass.

$\begin{array}{rcl}{m}_2& =& 180 \mathrm{g}\times {\left(\frac{{\displaystyle 50 \mathrm{Hz}}}{{\displaystyle 30 \mathrm{Hz}}}\right)}^2\\ & =& 500 \mathrm{g}\end{array}$