A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time:
A
at the highest position of the platform
B
at the mean position of the platform
C
for an amplitude of 
D
for an amplitude of 
Text Solution
Verified by ExpertsThe correct answer is:
C
As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they doesn't get separated) increases.
If we draw the FBD for coin at one of the extreme positions as shown
then from Newton's law, 
For loosing contact with the platform, 
So, 
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…