Question

A sonometer wire of length $1.5\text{m}$ is made of steel. The tension in it produces an elastic strain of $1\%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7\times {10}^3{\text{kg/m}}^3$ and $2.2\times {10}^{11}{\text{N/m}}^2$ respectively?

• A. $188.5\text{Hz}$
• B. $178.2\text{Hz}$
• C. $770\text{Hz}$
• D. $200.5\text{Hz}$

Answer 1

The correct option is B $178.2\text{Hz}$

Frequency, $f=\frac{v}{2l}=\frac{1}{2l}\sqrt{\frac{T}{\mu }}=\frac{1}{2l}\sqrt{\frac{T}{A\rho }}$

Where, $T=$ tension in the wire and $\mu =$ mass per unit length of wire

Also, Young's modulus, $Y=\frac{Tl}{A\Delta l}$

$\Rightarrow \frac{T}{A}=\frac{Y\Delta l}{l}$

Putting the value in expression of frequency,

$f=\frac{1}{2l}\sqrt{\frac{Y\Delta l}{ld}}$ .......(1)

Given:$l=1.5\text{m},\frac{\Delta l}{l}=0.01,\rho =7.7\times {10}^3{\text{kg/m}}^3$
and $y=2.2\times {10}^{11}{\text{N/m}}^2$

On putting these values in eq. (1),

$f=\frac{1}{2l}\sqrt{\frac{2.2\times {10}^{11}\times 0.01}{7.7\times {10}^3}}$

$f=\frac{{10}^3}{3}\sqrt{\frac{2}{7}}$

$f\approx 178.2\text{Hz}$

Hence, option $\left(B\right)$ is correct.