The bob of simple pendulum executes S.H.M in water with the a period t, while the period of oscillation of the bob is `t_(0)` in the air , neglecting frictional force of water and given that the density of the bob is `(4000)/(3)kgm^(-3)` , Find the relationship between `t and t_(0)` ?

Explanation

In air `t_(0)=2pisqrt((l)/(g))`
Let V be the volume of the bob. Then apparent weight of bob in water = weight of bob in air - up thrust
`Vrhog' =Vrhog-Vsigmag`
`cancel(V)rhog'=cancel(V)(rhog-sigmag)`
`g'=(g-(sigma)/(rho)g)`
`g'=g(1-(sigma)/(rho))`
`g'=(1-(sigma)/(rho))g`
Density of bob , `rho=(4000)/(3) kgm^(-3)`
Density of water , `sigma=1000kgm^(-3)`
`g'=(1-(1000xx3)/(4000))g`
`=g/4`
Time period of the pendulum in water
`t=2pisqrt((l)/(g'))=2pisqrt((l)/(g/(4)))=2xx2pisqrt((l)/(g'))`
`:.=t_(0)=2t.`