A sonometer wire is stretched by hanging a metal bob. The fundamental frequency of vibration of wire is ' n 1 '. When the bob is completely immersed in water, the frequency of vibration of

Question

A sonometer wire is stretched by hanging a metal bob. The fundamental frequency of vibration of wire is ' n 1 ​ '. When 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

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

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

The fundamental frequency in air is $n_{1} = \frac{1}{2 L} \frac{V ρ g}{μ}$ .

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.

The fundamental frequency in water is $n_{2} = \frac{1}{2 L} \frac{V ( ρ - ρ _{w} ) g}{μ}$ .

Taking the ratio of the squares of the frequencies:

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

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

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

$n_{1}^{2} (R D - 1) = n_{2}^{2} R D$

$R D (n_{1}^{2} - n_{2}^{2}) = n_{1}^{2}$

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

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

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