Question
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 $${t_0}$$ in air. Neglecting frictional force of water and given that the density of the bob is $$\left( {\frac{4}{3}} \right) \times 1000kg/{m^3}.$$ What relationship between $$t$$ and $${t_0}$$ is true
A.
$$t = 2{t_0}$$
B.
$$t = \frac{{{t_0}}}{2}$$
C.
$$t = {t_0}$$
D.
$$t = 4{t_0}$$
Answer :
$$t = 2{t_0}$$
Solution :
$$t = 2\pi \sqrt {\frac{\ell }{{{g_{eff}}}}} ;\,{t_0} = 2\pi \sqrt {\frac{\ell }{g}} $$
$$\eqalign{ & m{g_{eff}} = mg - B = my - V \times 100 \times g \cr & \therefore {g_{eff}} = g - \frac{{100}}{{\left( {\frac{m}{v}} \right)}}g = g - \frac{{1000}}{{\frac{4}{3} \times 1000}}g = \frac{g}{4} \cr & \therefore t = 2\pi \sqrt {\frac{\ell }{{\frac{g}{4}}}} \cr & \,t = 2{t_0} \cr} $$
$$t = 2\pi \sqrt {\frac{\ell }{{{g_{eff}}}}} ;\,{t_0} = 2\pi \sqrt {\frac{\ell }{g}} $$
$$\eqalign{ & m{g_{eff}} = mg - B = my - V \times 100 \times g \cr & \therefore {g_{eff}} = g - \frac{{100}}{{\left( {\frac{m}{v}} \right)}}g = g - \frac{{1000}}{{\frac{4}{3} \times 1000}}g = \frac{g}{4} \cr & \therefore t = 2\pi \sqrt {\frac{\ell }{{\frac{g}{4}}}} \cr & \,t = 2{t_0} \cr} $$