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. $250.5\text{Hz}$
• D. $770\text{Hz}$

Answer 1

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

Given:
Length of wire, $l=1.5\text{m}$
Elastic strain, $\frac{\Delta l}{l}=1\%=0.01$
Density, $\rho =7.7\times {10}^3{\text{kg/m}}^3$
Elasticity of steel, $Y=2.2\times {10}^{11}{\text{N/m}}^2$

Let $T$ be the tension in the wire and $A$ be the cross-sectional area.

Fundamental frequency of a sonometer wire is given by
$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}=\frac{1}{2l}\sqrt{\frac{T}{\rho A}}.....\left(1\right)$
Also, Young's modulus,
$Y=\frac{T/A}{\Delta l/l}$
$\Rightarrow \frac{T}{A}=\frac{Y\Delta l}{l}........\left(2\right)$

From Eq. (1) and (2),
$f=\frac{1}{2l}\sqrt{\frac{Y\Delta l}{\rho l}}$
$=\frac{1}{2\times 1.5}\sqrt{\frac{2.2\times {10}^{11}\times 0.01}{7.7\times {10}^3}}=178.2\text{Hz}$
Hence, option $\left(b\right)$ is true.