wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest. Then, the period of oscillations is (in seconds):
[Take g=10 m/s2]

A
2π5
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2π10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4π5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3π10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2π5
Let the mass of the block be m.
The block and plank are at momentary rest when the plank reaches the extreme position.
To lose contact, normal reaction = 0.
Taking the plank as a frame of reference, the block will face a pseudo force mamax in the upward direction at the top extreme position.


At extreme top position, applying Newton's 2nd law,
Fy=0
N+mamax=mg
N=0amax=g

We know, at extreme position, amax=Aω2
Aω2=gω=gA=100.4=25
or ω=5 rad/s

Time period of oscillation =2πω=2π5 s
option (a) is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon