wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 3.92×103 m. What should be the least period of these oscillations, so that the object is not detached from the platform ?

A
0.1256 sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0.1356 sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.1456 sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.1556 sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 0.1256 sec

From FBD of object,
mgR=ma
where, m= mass of the object,
R= Reaction force,
a= acceleration of the body,
ω= angular frequency of SHM.
For critical condition of object just losing contact with platform surface, R=0.
So, mg=ma
mg=mAω2
where A= amplitude of SHM [since Aω2 is the max. acceleration]
ω=g/A=9.83.92×103=50 rad/s
T=2πω=2π50=0.1256 sec
Hence, option (a) is the correct answer.

flag
Suggest Corrections
thumbs-up
12
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon