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 $(4/3)\times 1000kg/m^3$ . What relationship between $t$ and $t_0$ is true?

• A. $t=t_0$
• B. $t=t_0/2$
• C. $t=2t_0$
• D. $t=4t_0$

Answer 1

Answer: (C)

The time period of simple pendulum in air $T=t_{.0}=2\pi \sqrt{\left(\frac{l}{g}\right)}$ ?. (i) $l$ ,being the length of simple pendulum. In water, effective weight of bob $w=$ weight of bob in air - upthrust $\Rightarrow$ $\rho V{g}_{eff}=mg-mg$ $=\rho Vg-\rho Vg=(\rho -\rho )Vg$ where $\rho =$ density of bob, $\rho =$ density of water $\therefore$ $g_{eff}=\left(\frac{\rho -\rho }{\rho }\right)g=\left(1-\frac{\rho }{\rho }\right)g$ $\therefore$ $t=2\pi \sqrt{\left[\frac{l}{\left(1-\frac{\rho }{\rho }\right)g}\right]}$ ? (ii) Thus, $\frac{t}{t_0}=\sqrt{\left(\frac{1}{\left(1-\frac{\rho }{\rho }\right)}\right)}$ $=\sqrt{\left(\frac{1}{1-\frac{1000}{(4/3)\times 1000}}\right)}=\sqrt{\left(\frac{4}{4-3}\right)}$ $=2\Rightarrow t=2t_0$