# The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t in air. Neglecting frictional force of water and given that the density of the bob is 3 kg/m3. What relationship between t and t0 is true?

Saurabh Koranglekar
2 years ago
Arun
25758 Points
2 years ago

As we learnt in

Time period of oscillation of simple pendulum -

$T=2\pi \sqrt{\frac{l}{g}}$

- wherein

l = length of pendulum

g = acceleration due to gravity.

$\dpi{100} t_{0}=2\pi \sqrt{l/g}.........(i)$

Due to upthrust of water on the top, its apparent weight decreases

upthrust = weight of liquid displaced

$\dpi{100} \therefore \; \; \; Effective\; weight=mg-(V\sigma g)=V\rho g-V\sigma g$

$\dpi{100} V\rho g'=Vg(\rho -\sigma ),where\; \sigma \; is \; density \; of\; water\;$

$\dpi{100} or\; \; g'=g\left ( \frac{\rho -\sigma }{\rho } \right )$

$\dpi{100} \therefore \; \; t=2\pi \sqrt{l/g'}=2\pi \sqrt{\frac{l\rho }{g(\rho -\sigma )}}..........(ii)$

$\dpi{100} \therefore \; \; \; \frac{t}{t_{0}}=\sqrt{\frac{l\rho }{g(\rho -\sigma )}\times \frac{g}{l}}=\sqrt{\frac{\rho }{\rho -\sigma }}=\sqrt{\frac{4\times 1000/3}{(\frac{4000}{3}-1000)}}=2$

$\dpi{100} or\; \; \; t=t_{0}\times 2=2t_{0}$