The bob of simple pendulum performs S.H.M. with a frequency ′ n ′ in air and with frequency ′ n 1 ′ in water. If the density of the material of bob is ′ x ′ g / cc and the ratio n / n 1 is 3, then

Question

The bob of simple pendulum performs S.H.M. with a frequency ′ n ′ in air and with frequency ′ n 1 ′ ​ in water. If the density of the material of bob is ′ x ′ g / cc and the ratio n / n 1 ​ is 3, then density of bob ′ x ′ is ( density of water = 1 g /cc and neglect friction due to water)

TestHub · Accepted Answer

The frequency of a simple pendulum in air is given by $n = \frac{1}{2 π} \frac{g}{l}$ .

When the bob is immersed in water, the effective acceleration due to gravity becomes $g^{'} = g (1 - \frac{ρ _{w a t er}}{ρ _{b o b}})$ .

Given $ρ_{w a t er} = 1$ g/cc and $ρ_{b o b} = x$ g/cc, we have $g^{'} = g (1 - \frac{1}{x}) = g (\frac{x - 1}{x})$ .

The frequency of the pendulum in water is $n_{1} = \frac{1}{2 π} \frac{g ^{'}}{l}$ .

The ratio of the frequencies is $\frac{n}{n _{1}} = \frac{g}{g ^{'}} = \frac{x}{x - 1}$ .

Given $\frac{n}{n _{1}} = 3$ , we get $\frac{x}{x - 1} = 3$ .

Squaring both sides, $\frac{x}{x - 1} = 9$ .

$x = 9 x - 9$

$8 x = 9 \Rightarrow x = \frac{9}{8}$ g/cc.

Answer: $\frac{9}{8}$

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