Question

A wire of density $9\times {10}^{-3}kgc{m}^{-3}$ is stretched between two clamps $1m$ 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 $Y=9\times {10}^{10}N{m}^{-2}$), (to the nearest integer), _______.

Answer 1

Step 1. Given data.

Density is $\rho =9\times {10}^{-3}kgc{m}^{-3}$

Strain in the wire is $∆l=4.9\times {10}^{-4}.$

Young’s modulus of wire $Y=9\times {10}^{10}N{m}^{-2}$

Stretched length is $l=1m$

Step 2. calculating frequency

We know frequency relation is $f=\frac{1}{2l}\sqrt{\frac{Y∆l}{\rho l}}$

Where, $f$ is the frequency,

$Y$ is Young’s modulus of wire

$∆l$ is the Strain in the wire

$\rho$ is the density

$l$ is the Stretched length

By substituting the given values, we get

$$\begin{gathered} f=\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{9000\times 1}} \\ =35HZ \end{gathered}$$

Therefore, the lowest frequency of the transverse vibrations is $35Hz$.