SHMHard
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
The potential energy of a particle executing S.H.M. at a distance x from mean position is :...The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the tim...The circular motion of a particle with constant speed is...Two waves are represented by the equations y1 = a sin (ωt + kx + 0.57) m and y2 = a sin(ωt + kx)m, where x is ...A particle executing S.H.M. of amplitude 4 cm and T = 4 sec. The time taken by it to move from positive extreme position...