 The fundamental frequency of sonometer wire is $n$. If the length, tension and diameter are tripled, the new fundamental frequency is:\begin{align} & A.\ \ \ \dfrac{n}{\sqrt{3}} \\ & B.\ \ \ \dfrac{n}{3} \\ & C.\ \ \ n\sqrt{3} \\ & D.\ \ \ \dfrac{n}{3\sqrt{3}} \\ \end{align} Verified
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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
$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$
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 $A=\dfrac{\pi {{d}^{2}}}{4}$

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
$n=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$
If the wire has $\text{density}=\rho$. Then mass of the object is given by

\begin{align} & m=\text{ volume}\times \text{density} \\ & \Rightarrow m\text{=Area}\times \text{length}\times \text{density} \\ & \Rightarrow m=\pi \dfrac{{{d}^{2}}}{4}\times l\times \rho =\dfrac{\pi {{d}^{2}}l\rho }{4} \\ \end{align}

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

When the length, diameter and tension is tripled then the new frequency will be
$n'=\dfrac{1}{2l'}\sqrt{\dfrac{T'}{m'}}$
Here $l'=3l,T'=3T,$
The mass of the wire now will be
$m=\dfrac{\pi d{{'}^{2}}l'\rho }{4}=\dfrac{\pi {{\left( 3d \right)}^{2}}3l\rho }{4}=27\dfrac{\pi {{d}^{2}}l\rho }{4}=27m$ ($\because m=\dfrac{\pi {{d}^{2}}l\rho }{4}$)
Now the frequency becomes,
$n'=\dfrac{1}{2\times 3l}\sqrt{\dfrac{3T}{27m}}=\dfrac{1}{9}\times \dfrac{1}{2l}\sqrt{\dfrac{T}{m}}=\dfrac{n}{9}$

So, the correct answer is “Option B”.

$\text{Time period}=\dfrac{1}{\text{frequency}}$