The bob of simple pendulum performs S.H.M. with a frequency 'n' in air and with frequency 'n₁' in water. If the density of the material of bob is 'x' g / cc and the ratio n / n₁ is 3, then density of bob 'x' is ( density of water = 1 g /cc and neglect friction due to water)

Step-by-step solution

The frequency of a simple pendulum in air is given by n = 1 2 g l . When the bob is immersed in water, the effective acceleration due to gravity becomes g' = g ( 1 - _ water _ bob ) . Given _ water = 1 g/cc and _ bob = x g/cc, we have g' = g ( 1 - 1 x ) = g ( x - 1 x ) . The frequency of the pendulum in water is n₁ = 1 2 g' l . The ratio of the frequencies is n n₁ = g g' = x x - 1 . Given n n₁ = 3 , we get x x - 1 = 3 . Squaring both sides, x x - 1 = 9 . x = 9x - 9 8x = 9 x = 9 8 g/cc. Answer: 9 8