Question

The fundamental frequency of sonometer wire is $n$. If the length, tension and diameter are tripled, the new fundamental frequency is:
$$\begin{array}{rl}& A.{\displaystyle \frac{n}{\sqrt{3}}}\\ & B.{\displaystyle \frac{n}{3}}\\ & C.n\sqrt{3}\\ & D.{\displaystyle \frac{n}{3\sqrt{3}}}\end{array}$$

Answer 1

Hint: First calculate the frequency of sonometer wire at its original length, diameter and tension. After that calculate the frequency of sonometer wire when length, diameter, and tension is tripled. Compare both the frequency to get a relation between them.

Formula used:
Frequency of oscillation of any wire of length $l$ mass $m$ and tension on string $T$ is
$f={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}$
Mass of any material of volume $V$and density $\rho$is $m=V\rho$
Volume$V=\text{ length}\times \text{ area of cross-section}$
Area of cross-section of a surface of diameter $d$ is $A={\displaystyle \frac{\pi d^2}{4}}$

Complete answer:
Let the initial length of the wire be $l$,diameter be $d$ and the tension on the wire be $T$, then the frequency is given by
$$n={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}$$
If the wire has $\text{density}=\rho$. Then mass of the object is given by

$\begin{array}{rl}& m=\text{ volume}\times \text{density}\\ & \Rightarrow m\text{=Area}\times \text{length}\times \text{density}\\ & \Rightarrow m=\pi {\displaystyle \frac{d^2}{4}}\times l\times \rho ={\displaystyle \frac{\pi d^{2}l\rho }{4}}\end{array}$

 So, $$n={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{\left({\displaystyle \frac{\pi d^{2}l\rho }{4}}\right)}}}$$

When the length, diameter and tension is tripled then the new frequency will be
$$n^{\prime }={\displaystyle \frac{1}{2l^{\prime }}}\sqrt{{\displaystyle \frac{T^{\prime }}{m^{\prime }}}}$$
Here $l^{\prime }=3l,T^{\prime }=3T,$
The mass of the wire now will be
$m={\displaystyle \frac{\pi d{{}^{\prime }}^2{l}^{\prime }\rho }{4}}={\displaystyle \frac{\pi {\left(3d\right)}^{2}3l\rho }{4}}=27{\displaystyle \frac{\pi d^{2}l\rho }{4}}=27m$ ($\because m={\displaystyle \frac{\pi d^{2}l\rho }{4}}$)
Now the frequency becomes,
$$n^{\prime }={\displaystyle \frac{1}{2\times 3l}}\sqrt{{\displaystyle \frac{3T}{27m}}}={\displaystyle \frac{1}{9}}\times {\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}={\displaystyle \frac{n}{9}}$$

So, the correct answer is “Option B”.

Additional Information:
Frequency is defined as the number of oscillations or occurrence per unit time. The unit of frequency is Hertz. Frequency is also defined as the reciprocal of time period. From the above calculation you can see that by increasing the length and diameter of the cross-section of wire the frequency decreases.

Note:
For problems like this, first calculate frequency in the original case, and then calculate the frequency for the changed case. Compare both of them to get a relationship between them. You can also calculate the time period because the time period is reciprocal of frequency.

$\text{Time period}={\displaystyle \frac{1}{\text{frequency}}}$

When the tension on the wire is large the wire will break. This breaking point is called breaking stress. So breaking stress is defined as the force on the wire per unit cross-sectional area.