Question

A simple pendulum has a time period of T${}_0$ on the surface of the earth. On another planet whose density is same that of earth but radius is 4 times then time period of this simple pendulum would be :

• A. T${}_0$
• B. $\frac{T_0}{4}$
• C. $\frac{T_0}{2}$
• D. 4T${}_0$

Answer 1

Answer: (C)

$\frac{T_0}{2}$

Acceleration due to gravity on earth $g=\frac{GM}{R^2}$ where $M=\rho \times \frac{4\pi }{3}{R}^3$

$⟹$ $g=\frac{4}{3}\pi \rho GR=KR$

Acceleration due to gravity on other planet, $g^{\prime }=\frac{4}{3}\pi \rho G\left(R^{\prime }\right)=K\left(4R\right)=4g$

Time period on earth, $T_o=2\pi \sqrt{\frac{l}{g}}$

Time period on planet, $T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}=\frac{1}{2}\times 2\pi \sqrt{\frac{l}{g}}=\frac{T_o}{2}$