Question

Two simple pendulums of length 0.5 m and 2 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed x oscillations, where x is:

• A. 1
• B. 3
• C. 2
• D. 5

Answer 1

The correct option is C 2

Time period of oscillation of a simple pendulum is given by:

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

Thus, the ratio of their time periods is $\frac{T_1}{T_2}=\sqrt{\frac{0.5}{2}}=\frac{1}{2}$

$⟹T_2=2T_1$

Thus, the shorter pendulum would have completed exactly 2 oscillations when the longer pendulum would have completed the first oscillation.