Question

The fundamental frequency that can be excited in a sonometer wire of length $L$ is $f$. If the maximum stress this wire can bear without exceeding the elastic limit is $S$, the density of the wire is :

• A. $\frac{S}{2L^2{f}^2}$
• B. $\frac{S}{4L^2{f}^2}$
• C. $\frac{4S}{L^2{f}^2}$
• D. $\frac{2S}{L^2{f}^2}$

Answer 1

The correct option is B $\frac{S}{4L^2{f}^2}$

Given maximum stress is $S$
$S=\frac{F}{A}\Rightarrow F=SA.$
Fundamental frequency
$f=\frac{1}{2L}\sqrt{\frac{T}{m}}$ Where $T=$Tension in wire$=F$ and $m=$mass/length$=M/L$

$\therefore f=\frac{1}{2L}\sqrt{\frac{F}{M/L}}=\frac{1}{2L}\sqrt{\frac{SA}{M/L}}=\frac{1}{2L}\sqrt{\frac{S}{M/AL}}$

But $\frac{M}{AL}=\rho =$density of the wire

$\therefore f=\frac{1}{2L}\sqrt{\frac{S}{\rho }}$

$\therefore \rho =\frac{S}{4L^2{f}^2}=$Density of the wire