A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude form the equilibrium position is:
A
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
C
Let displacement equation of particle executing SHM is
y = a sin ꞷt
As particle travels half of the amplitude from the equilibrium position, so
y = 
Therefore, 
= a sin ꞷt
or sin ꞷt = 
= sin 
or ꞷt = 
or t = 
or t = 
or t = 
Hence, the particles travels half of he amplitude from the equilibrium in 
sec.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The equation of S.H.M. is y = a sin (2 π nt + α), then its phase at time is
Two equation of two S.H.M. are x = a sin (ꞷt – α) and y = b cos(ꞷt – α). The phase difference betwe…
A particle is oscillating according to the equation X = 7 cos 0.5 π t, where t is in second. The po…
The amplitude and the periodic time of a S.H.M. are 5 cm and 6 sec respectively. At a distance of 2…
A 1.00 x 10 –20 kg particle is vibrating with simple harmonic motion with a period of 1.00 x 10 –5 …
The displacement of a particle moving in S.H.M. at any instant is given by y = a sin ꞷt. The accele…