Question

A sonometer wire of length$1.5m$ is made of steel. The tension in it produces an elastic strain of $1\%$. What is the fundamental frequency of steel if the density and elasticity of steel are $7.7\times {10}^{3}kg/m^3$ and $2.2\times {10}^{11}N/m^2$ respectively?

• A. $188.5Hz$
• B. $178.2Hz$
• C. $200.5Hz$
• D. $770Hz$

Answer 1

The correct option is B

$178.2Hz$

Step 1 : Given data

Length $l=1.5m$

Elastic strain $=1\%$

Density of steel $=7.7\times {10}^{3}kg/m^3$

Elasticity of steel $=2.2\times {10}^{11}N/m^3$

Step 2: To find the fundamental frequency of steel

$$\begin{gathered} f=\frac{v}{2l} \\ =\frac{1}{2l}\sqrt{\frac{T}{\mu }} \\ =\frac{1}{2l}\sqrt{\frac{T}{Ad}} \end{gathered}$$

Where$T$ is tension in the wire and $\mu$ is the mass per unit length of wire

Young's modulus

$$\begin{gathered} Y=\frac{Tl}{A∆l} \\ \frac{T}{A}=\frac{Y∆l}{l} \end{gathered}$$

Therefore frequency , $f=\frac{1}{2l}\sqrt{\frac{Y∆l}{ld}}$

Putting the given values

$$\begin{gathered} f=\frac{1}{2l}\sqrt{\frac{2.2\times {10}^{11}\times 0.01}{7.7\times {10}^3}} \\ =\sqrt{\frac{2}{7}}\times \frac{{10}^3}{3} \\ f\approx 178.2Hz \end{gathered}$$

Hence the fundamental frequency of steel $=178.2Hz$

Therefore the correct answer is Option B