Question

A string $1.5\text{m}$ long is made of steel of density $8400kg/m^3$ and $Y=9\times {10}^{11}N/m^2$, wire is maintained at tension which produces strain of $2.1\times {10}^{-2}$ in string, the fundamental frequency of transverse of string is _____. (String is clamped at both ends)

• A. $400\text{Hz}$
• B. $500\text{Hz}$
• C. $600\text{Hz}$
• D. $900\text{Hz}$

Answer 1

The correct option is B $500\text{Hz}$

We know that,
$Y=\frac{stress}{strain}$

$\frac{F/A}{\Delta l/l}=Y$
$T=YA\left(\frac{\Delta l}{l}\right)$

Velocity of wave along a stretched string is given by:
$v=\sqrt{T/\mu }$
Here, $\mu =$ mass per unit length of the string
$v=\sqrt{\frac{YA\Delta l/l}{m/l}}v=\sqrt{Y\Delta l/l\left(\frac{V}{m}\right)}$
$v=\sqrt{\frac{Y}{\rho }\frac{\Delta l}{l}}$
$v=\sqrt{\frac{9\times {10}^{11}\times 2.1}{8400\times 100}}v=\sqrt{\frac{9\times {10}^{10}\times 21}{84\times {10}^4}}$

$\Rightarrow v=\frac{3}{2}\times {10}^3=1500\text{m/s}$
$n=\frac{v}{2l}=\frac{1500}{2\times 1.5}=500\text{Hz}$