Q141 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Physics
A steel wire of length $1.25 \mathrm{~m}$ is stretched between two rigid supports. The tension in the wire produces an elastic strain of $0.14 \%$. The fundamental frequency of the wire is (Density and Young's modulus of steel are $7.7 \times 10^3 \mathrm{kgm}^{-3}$ and $2.2 \times 10^{11} \mathrm{Nm}^{-2}$ respectively)
- A$20 \mathrm{~Hz}$
- B$40 \mathrm{~Hz}$
- Correct answerC$80 \mathrm{~Hz}$
- D$160 \mathrm{~Hz}$
Solution
Given, length of wire $=1.25 \mathrm{~m}$
strain produced $=0.14 \%$
density of wire, $d=7.7 \times 10^3 \mathrm{kgm}^{-3}$
young's modulus, $Y=2.2 \times 10^{11} \mathrm{Nm}^{-2}$
Fundamental frequency of vibration over a string is given by
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}
$$
Where, $T=$ Tension in string,
$L=$ Length of string and
$\mu=$ mass per unit length of string.
So, we can express frequency as
$$
f=\frac{1}{2 L} \sqrt{\frac{T / A}{M / l A}}
$$
From young's modulus of elasticity we have,
$$
\begin{aligned}
& & Y=\frac{T / A}{\Delta l / l} \\
\Rightarrow & \frac{T}{A} & =2.2 \times 10^{11} \times \frac{0.14}{100} \\
\Rightarrow & \frac{T}{A} & =0.308 \times 10^9
\end{aligned}
$$
Substituting for $L, \frac{T}{A}$ and $d$ in eq. (i) we get, Fundamental frequency,
$$
\begin{aligned}
f & =\frac{1}{2 \times 1.25} \times \sqrt{\frac{0.308 \times 10^9}{7.7 \times 10^3}} \\
& =\frac{1}{25} \times \sqrt{0.04 \times 10^6} \\
& =\frac{1}{25} \times \sqrt{4 \times 10^4}=\frac{2}{25} \times 10^2=80 \mathrm{~Hz}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.