In a sonometer experiment, the fundamental frequency of vibration of wire is ' n 1 ' when wire is stretched by hanging a metal bob. If the bob is completely immersed in water, the frequenc

Question

In a sonometer experiment, the fundamental frequency of vibration of wire is ' n 1 ​ ' when wire is stretched by hanging a metal bob. If the bob is completely immersed in water, the frequency of vibration of wire becomes ' n 2 ​ '. The relative density of the metal of the bob is

TestHub · Accepted Answer

The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2 L} \frac{T}{m}$ , where $T$ is the tension and $m$ is the linear mass density.

When the bob is in air, the tension is $T_{1} = V ρ g$ , where $V$ is the volume and $ρ$ is the density of the bob.

$n_{1} = \frac{1}{2 L} \frac{V ρ g}{m}$

When the bob is completely immersed in water, the tension becomes $T_{2} = V ρ g - V ρ_{w} g = V (ρ - ρ_{w}) g$ , where $ρ_{w}$ is the density of water.

$n_{2} = \frac{1}{2 L} \frac{V ( ρ - ρ _{w} ) g}{m}$

Squaring and taking the ratio of the frequencies:

$\frac{n _{1}^{2}}{n _{2}^{2}} = \frac{T _{1}}{T _{2}} = \frac{V ρ g}{V ( ρ - ρ _{w} ) g} = \frac{ρ}{ρ - ρ _{w}}$

Dividing the numerator and denominator by $ρ_{w}$ , we get the relative density $d = \frac{ρ}{ρ _{w}}$ :

$\frac{n _{1}^{2}}{n _{2}^{2}} = \frac{d}{d - 1}$

$n_{1}^{2} d - n_{1}^{2} = n_{2}^{2} d$

$d (n_{1}^{2} - n_{2}^{2}) = n_{1}^{2}$

$d = \frac{n _{1}^{2}}{n _{1}^{2} - n _{2}^{2}}$

Answer: $\frac{n _{1}^{2}}{n _{1}^{2} - n _{2}^{2}}$

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion