JEE Main 2023 — Physics Waves & Oscillations

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

$$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}...(1)$$f=2L1​μT​​...(1)

For the first mass, it can be written that

$${f}_{1}=\frac{1}{2L}\sqrt{\frac{{m}_{1}g}{\mu }}...(2)$$f1​=2L1​μm1​g​​...(2)

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

$${f}_{2}=\frac{1}{2L}\sqrt{\frac{{m}_{2}g}{\mu }}...(3)$$f2​=2L1​μm2​g​​...(3)

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

$$\begin{matrix}\frac{{f}_{2}}{{f}_{1}} & = & \frac{\frac{1}{2L}\sqrt{\frac{{m}_{2}g}{\mu }}}{\frac{1}{2L}\sqrt{\frac{{m}_{1}g}{\mu }}} \\ & = & \sqrt{\frac{{m}_{2}}{{m}_{1}}} \\ & \Rightarrow & \frac{{m}_{1}}{{m}_{2}}={(\frac{{f}_{1}}{{f}_{2}})}^{2} \\ & \Rightarrow & {m}_{2}={m}_{1}{(\frac{{f}_{2}}{{f}_{1}})}^{2}...(4)\end{matrix}$$f1​f2​​​==⇒⇒​2L1​μm1​g​​2L1​μm2​g​​​m1​m2​​​m2​m1​​=(f2​f1​​)2m2​=m1​(f1​f2​​)2...(4)​

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

$$\begin{matrix}{m}_{2} & = & 180g\times {(\frac{50\mathrm{Hz}}{30\mathrm{Hz}})}^{2} \\ & = & 500g\end{matrix}$$m2​​==​180g×(30Hz50Hz​)2500g​