Question

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacements in one




direction at the same time. They will again be in the same phase when the pendulum of shorter length has




completed ______ oscillations.




(1) 2 (2) 1




(3) 5 (

Answer 1

Answer:

As 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.

 

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

N_{S} 2\pi \sqrt{\frac{5}{g}} = N_{L}\times 2\pi\sqrt{\frac{20}{g}}

\therefore N_{S} = 2N_{L}\ \ \ \ where \ \ N_{L} = 1

\therefore N_{S} = 2

Explanation: