Question

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of $1\%$ . What is the fundamental frequency of steel if 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.)178.2 Hz
B.)200.5 Hz
C.)770 Hz
D.)188.5 Hz

Answer 1

Hint: to find the solution to this problem obtain the mathematical expression for fundamental frequency. Express the constituents of the fundamental frequency in terms of the given quantities in the question. Then we can put the value to find the required solution.

Complete step by step answer:
The length of the sonometer wire which is made of steel is $l=1.5m$
The tension in the wire produces $1\%$ of elastic strain.
The density of the steel or the wire is, $\rho =~7.7\times {{10}^{3}}kg/{{m}^{3}}$
The elasticity of the wire is, $Y=2.2\times {{10}^{11}}N/{{m}^{2}}$

Now we need to find the fundamental frequency of the wire.

The fundamental frequency of the can be mathematically expressed as, $\nu =\dfrac{v}{2l}$
Again, the velocity in the wire is given by, $v=\sqrt{\dfrac{T}{\mu }}$
Where T is the tension on the wire and $\mu$ is the mass per unit length of the wire.

$\mu =\dfrac{m}{l}$

Putting these values, we can write,

$\nu =\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

Again, if A is the cross-sectional area of the wire and l is the length, then the volume of the wire will be, $\text{volume }=\text{ }A.l$

Again, the density of the wire is given as mass per unit volume. So,

$\begin{aligned} & \text{density = }\dfrac{\text{mass}}{\text{volume}} \\\ & \rho =\dfrac{m}{Al} \\\ & m=\rho Al \\\ \end{aligned}$

Putting this on the equation for fundamental frequency we get that,

$\begin{aligned} & \nu =\dfrac{1}{2l}\sqrt{\dfrac{Tl}{\rho Al}} \\\ & \nu =\dfrac{1}{2l}\sqrt{\dfrac{T}{\rho A}} \\\ \end{aligned}$

Now, elasticity of the wire is given by,

$\begin{aligned} & \text{elasticity = }\dfrac{\text{stress}}{\text{longitudinal strain}} \\\ & Y=\dfrac{\dfrac{T}{A}}{\dfrac{\Delta l}{l}} \\\ & Y=\dfrac{Tl}{A\Delta l} \\\ & \dfrac{T}{A}=\dfrac{Y\Delta l}{l} \\\ \end{aligned}$

So, we can write,

$\nu =\dfrac{1}{2l}\sqrt{\dfrac{Y}{\rho }\dfrac{\Delta l}{l}}$
Again, we know that, $\dfrac{\Delta l}{l}$ is the strain. The wire produces a strain of $1\%$ . So,

$\dfrac{\Delta l}{l}=1\%=\dfrac{1}{100}=0.01$

Putting the values on the above equation, we get that,

$\begin{aligned} & \nu =\dfrac{1}{2l}\sqrt{\dfrac{Y}{\rho }\dfrac{\Delta l}{l}} \\\ & \nu =\dfrac{1}{2\times 1.5}\sqrt{\dfrac{2.2\times {{10}^{11}}}{7.7\times {{10}^{3}}}\times 0.01} \\\ & \nu =178.2Hz \\\ \end{aligned}$

The correct option is (A).

Note: Sonometer is a device which uses the principle of resonance. It is used to measure the frequency or the density of vibration. They are used in some medical processes to measure the bone density.