© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the fundamental frequency formula
The fundamental frequency of a vibrating string of length $ \ell $, tension $ T $, and linear mass density $ \mu $ is:
$ f = \frac{1}{2\ell} \sqrt{\frac{T}{\mu}} $
Alternatively, since $ \mu = \rho \, A $ (mass per unit length in terms of density $ \rho $ and cross-sectional area $ A $):
$ f = \frac{1}{2\ell} \sqrt{\frac{T}{\rho \, A}}. $
Step 2: Express tension in terms of Young’s modulus and strain
The problem states that the wire is stretched to produce an elastic strain of $1\%$, i.e.,
$ \frac{\Delta \ell}{\ell} = 0.01 $. The tension $ T $ can be related to Young’s modulus $ Y $ by:
$ Y = \frac{T \, \ell}{A \, \Delta \ell}
\quad \Rightarrow \quad
\frac{T}{A} = \frac{Y \, \Delta \ell}{\ell}.
$
Step 3: Substitute into the frequency formula
Replacing $ \frac{T}{A} $ with $ \frac{Y \, \Delta \ell}{\ell} $ and $ \mu $ with $ \rho \, A $, the fundamental frequency becomes:
$ f = \frac{1}{2 \ell} \sqrt{\frac{Y \, \Delta \ell}{\ell \, \rho}}.
$
Step 4: Plug in the values
Given data:
Wire length, $ \ell = 1.5 \, \text{m}$.
Strain, $ \frac{\Delta \ell}{\ell} = 0.01$.
Density of steel, $ \rho = 7.7 \times 10^3 \, \text{kg/m}^3$.
Young’s modulus of steel, $ Y = 2.2 \times 10^{11} \, \text{N/m}^2$.
Substituting into the formula:
$ f
= \frac{1}{2 \times 1.5}
\sqrt{\frac{2.2 \times 10^{11} \times (0.01 \times 1.5)}{1.5 \times 7.7 \times 10^3}}.
$
Upon careful calculation, this evaluates approximately to:
$ f \approx 178.2 \, \text{Hz}.
$
Step 5: Conclude the correct option
The fundamental frequency $ f $ is around $ 178.2 \, \text{Hz} $, which corresponds to the correct answer.
