The time period of a bob performing simple harmonic motion in water is 2   s . If density of bob is 4 3 × 10 3   kg   m - 3 , then time period of bob performing simple harmonic motion in air will be

Step-by-step solution

Given, density of bob, ρ = 4 3 × 10 3   kg   m - 3 Density of water, σ = 10 3     kg   m - 3 If g ' be gravitational acceleration in water then, g ' = g   1 - σ ρ = g   1 - 10 3 4 3 × 10 3 = g 4 As,    T air   = 2 π I g , Similarly,    T water   = 2 π I g '   T water   = 2   s   g ' = g 4 2 = 2 π   I g / 4 = 2 π   I g   · 2 2 = 2 . T air   ⇒