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? Density and Elasticity of steel are 7.7×${10^{ - 3}}$ $\dfrac{{Kg}}{{{m^3}}}$ and 2.2×${10^{11}}$ $\dfrac{N}{{{m^2}}}$ respectively?
A) 178.2 Hz
B) 200.5 Hz
C) 770 Hz
D) 188.5 Hz

Answer 1

We can simply start with the equation of the frequency and equating the values of tension as a force and $\mu$ as mass per unit length. After the young’s modulus, we can find tension per unit area and equate it in frequency value and get our result.

Formula used:
The magnetic field at a point O for a finite wire carrying a current $I$ is given by, $B = \dfrac{{{\mu _0}I}}{{4\pi d}}\left( {\sin {\theta _1} + {\theta _2}} \right)$ where $d$ is the perpendicular distance from the point O to the wire, ${\mu _0}$ is the permeability of free space and ${\theta _1}$, ${\theta _2}$ are the angles formed at point O by line segments joining each end to O.

Complete step by step answer:
Step 1:
First, we need to know about the fundamental frequency then we will proceed to the question:
The velocity of a traveling wave in a stretched string is determined by the tension and the mass per unit length of the string. for a string of length cm and mass per length = $\dfrac{{gm}}{m}$. For such a string, the fundamental frequency would be Hz. Any of the highlighted quantities can be calculated by clicking on them

Step 2:
We are given, A Sonometer wire of length $l= 1.5 m$ is made of steel, the tension $T$ in it produces an elastic strain or $\dfrac{{\Delta l}}{l}$ of 1%, and elasticity of steel ($Y$) and density $\rho$ also.
Formula for fundamental frequency is $\nu$ =$\dfrac{v}{{2l}}$, where v is the velocity and $l$ is length.
Here velocity can rewritten as $\sqrt {\dfrac{T}{\mu }}$ where $T$ is tension and $\mu$ is mass per unit length
So $\nu$ =$\dfrac{1}{{2l}}$ $\sqrt {\dfrac{T}{\mu }}$ or $\dfrac{1}{{2l}}$ $\sqrt {\dfrac{{Tl}}{m}}$, because $\mu$ is mass per unit length………………..… (1)
Now the formula for density is mass per unit volume then the mass is equal to $\rho V$ where V is volume
Then it can be rewritten as, m=$\rho Al$ $\Rightarrow$ $\dfrac{1}{{2l}}\sqrt {\dfrac{{Tl}}{{\rho Al}}}$ and length will get cancel out.
Now we know that: Y=$\dfrac{F}{A} \times \dfrac{l}{{\Delta l}}$ and here Tension is the force. So we can write this as
Y=$\dfrac{T}{A} \times \dfrac{l}{{\Delta l}}$, from here we will get the value of $\dfrac{T}{A}$
$\dfrac{T}{A}$ =$\dfrac{{Y\Delta l}}{l}$ ………………….. (2)
Putting the value of $\dfrac{T}{A}$ in (1)
$\Rightarrow$ $\nu$ =$\dfrac{1}{{2l}}$ $\sqrt {\dfrac{{Y\Delta l}}{{\rho l}}}$ here, again length inside the root gets cancel out and we will put all the given values stated above.
$\Rightarrow$ $\nu$ =$\dfrac{1}{{2 \times 1.5}}\sqrt {\dfrac{{2.2 \times {{10}^{11}} \times 0.01}}{{7.7 \times {{10}^3}}}}$ which will give value of frequency as 178.16Hz.

Hence, the fundamental frequency is 178.16Hz. Therefore, Option A is correct.

Note:
The Young's Modulus of a material is a fundamental property of every material that cannot be changed. It is dependent upon temperature and pressure. However, the Young's Modulus (or Elastic Modulus) is in essence the stiffness of a material. In other words, it is how easy it is bent or stretched.