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Question:

A wire of density $9 \times 10^{-3} \mathrm{~kg} \mathrm{~cm}^{-3}$ is stretched between two clamps $1 \mathrm{~m}$ apart. The resulting strain in the wire is $4.9 \times 10^{-4}$. The lowest frequency of the transverse vibrations in the wire is

(Young's modulus of wire $Y=9 \times 10^{10} \mathrm{Nm}^{-2}$ ), (to the nearest integer),

Solution:

$(35.00)$

Given, Denisty of wire, $\sigma=9 \times 10^{-3} \mathrm{~kg} \mathrm{~cm}^{-3}$

Young's modulus of wire, $Y=9 \times 10^{10} \mathrm{Nm}^{-2}$

Strain $=4.9 \times 10^{-4}$

$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\text { Strain }}$

$\therefore \frac{T}{A}=Y \times$ Strain $=9 \times 10^{9} \times 4.9 \times 10^{-4}$

Also, mass of wire, $m=A l \sigma$

Mass per unit length, $\mu=\frac{m}{I}=A \sigma$

Fundamental frequency in the string

$f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{2 l} \sqrt{\frac{T}{\sigma A}}=\frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{9} \times 4.9 \times 10^{-4}}{9 \times 10^{3}}}$

$=\frac{1}{2} \sqrt{49 \times 10^{9-4-3}}=\frac{1}{2} \times 70=35 \mathrm{~Hz}$