SHMMedium
Question
A simple pendulum of length 1 m is allowed to oscillated with amplitude 2o. It collides elastically with a wall inclined at 1o to the vertical. Its time period will be : (use g = π2)
Options
A.2/3 sec
B.4/3 sec
C.2 sec
D.None of these
Solution
Time period for half part : T = 
= 2sec.
So 2o part will be cornered in a time t =
= 1 sec.
For the left 1o part : θ = θ0 sin (ωt)1o = 2o sin
= π × t ⇒ t = 1/6 sec.
Total time =
+ 2t ⇒ 1 + 2 × 
= 1 + 
sec.
= 2sec.So 2o part will be cornered in a time t =
= 1 sec.For the left 1o part : θ = θ0 sin (ωt)1o = 2o sin
= π × t ⇒ t = 1/6 sec.Total time =
+ 2t ⇒ 1 + 2 × = 1 + sec.Create a free account to view solution
View Solution FreeMore SHM Questions
Time period of small oscillation (in a vertical plane normal to the plane of strings) of the bob in the arrangement show...A particle performing SHM takes time equal to T (time period of SHM) in consecutive appearances at a particular point. T...An object of mass m is suspended from a spring and it executes S.H.M. with frequency υ. If the mass is increased 4...A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass...A body executing SHM passes through its equilibrium. At this instant, it has...