Question

The fundamental frequency of a sonometer wire is $n$. If the length and diameter of the wire are doubled keeping the tension the same, then the new fundamental frequency is

• A. $\frac{2n}{\sqrt{2}}$
• B. $\frac{n}{2\sqrt{2}}$
• C. $\sqrt{2}n$
• D. $\frac{n}{4}$

Answer 1

The correct option is D

$\frac{n}{4}$

Step1: Given data and assumption.

The fundamental frequency of a sonometer wire$=n$

Let the length of the wire be $l$.

And the diameter of the wire be $d$

Step2: Find the new fundamental frequency when the length and diameter of the wire are doubled.

Formula used:

$f=\frac{1}{2l}\sqrt{\frac{T}{m}}$

Where,

$f$ is the fundamental frequency.

$l$ is the length of the wire.

$T$ is the tension of the wire.

$m$ is the mass per unit length.

Also,

Mass per unit length, $m=\frac{\rho v}{l}$

Where, $\rho$ is the density of the material of the wire, $v$ is the volume, and $l$ is the length of the wire.

Now,

$f=n=\frac{1}{2l}\sqrt{\frac{T}{m}}$ $\left[\because Given,f=n\right]$

$\Rightarrow$$n=\frac{1}{2l}\sqrt{\frac{T}{{\displaystyle \raisebox{1ex}{$\rho v$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}}$ $\left[\because m=\raisebox{1ex}{$\rho v$}\!\left/ \!\raisebox{-1ex}{$l$}\right.\right]$

$\Rightarrow$$n=\frac{1}{2l}\sqrt{\frac{T}{{\displaystyle \raisebox{1ex}{$\rho \pi r^{2}l$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}}$ $\left[\because v=\pi r^{2}l\right]$

$\Rightarrow$$n=\frac{1}{2l}\sqrt{\frac{T}{{\displaystyle \rho \pi r^2}}}$ $\left(i\right)$

Since if the length of the wire and diameter of the wire is double, the new frequency is.

$\Rightarrow$$n\text{'}=\frac{1}{2\left(2l\right)}\sqrt{\frac{T}{{\displaystyle \rho \pi {\left(2r\right)}^2}}}$

Where $n\text{'}$ is the new fundamental frequency.

$\Rightarrow$$n\text{'}=\frac{1}{4}\left(\frac{1}{2l}\sqrt{\frac{T}{{\displaystyle \rho \pi r^2}}}\right)$

$\Rightarrow$$n\text{'}=\frac{1}{4}n$ $\left[\because fromequation\left(i\right)\right]$

Hence, option D is correct.