Two simple pendulums of length 0.5 m and 20 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

Step-by-step solution

Time period of simple pendulum For Ist pendulum aligned T₁ & =2 5 g & =2 5 9.8 = 20 ~s aligned For IInd pendulum T₂=2 20 g =2 20 9.8 =2 20 ~s Suppose the time will be in same phase is t then aligned 1 t & = 1 T₁ - 1 T₂ & = 1 20 - 1 2 20 aligned = 1 2 20 So, t=2 20 Hence, the number of oscillation of simple pendulum of shorter length (x)= 2 20 20 =2 .