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.

The period of oscillation of a simple pendulum is 1.45 s. If the densi

1.	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.
Answer» Solution :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.