Questions
Two simple pendulums of length 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 .... oscillations.
detailed solution
Correct option is C
If t is the time taken by pendulums to come in same phase again first time after t=0.and NS=Number of oscillations made by shorter length pendulum with time period TS.NL= Number of oscillations made by longer length pendulum with time period TL.Then t=NSTS=NLTL⇒NS2π5g=NL×2π20g (∵T=2πlg)⇒NS=2NL i.e. if NL=1⇒NS=2Similar Questions
A simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in horizontal position from the wall and is left free to move. The bob hits on the wall the coefficient of restitution is . After how many collisions the amplitude of vibration will become less than 60°
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