The period of oscillation of a simple pendulum is 1.45 s. If the density of the material of the bob is `7.8xx10^(3)" kgm"^(-3)` and that of water is `1.0xx10^(3)" kgm"^(-3),` then calculate the period of oscillation of the simple pendulum in water.

Explanation

We know that `T.=2pisqrt((L)/(g(1-(rho)/(sigma))))`
`"i.e. "T=2pisqrt(((L)/(g))((sigma)/(sigma-rho)))" i.e "T.=Tsqrt((sigma)/(sigma-rho))`
Here`" "T=1.45 s, s=7.8xx10^(3)kgm^(-3)`
and `rho=1.0xx10^(3)kgm^(-3)`
Substituting the value we get,
`T=1.45 sqrt((7.8)/(7.8-1.0))xxsqrt((10^(3))/(10^(3)))=1.45xx1.071=1.55s.`
Hence the number of oscillation of a simple pendulum decreases in water.