CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The ends of a rod of length l and mass m are attached to two identical springs as shown in figure. The rod is free to rotate about its centre O. The rod is depressed slightly at end A and released. The time period of the resulting oscillation is


A
T=2π2mk
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
T=2πm2k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
T=π2m3k
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
T=2π3mk
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C T=π2m3k

Let the rod be depressed by a small distance x. Both the springs are compressed by x. When the rod is released, the restoring torque about O is given by
τ=(kx)×l2+(kx)×l2=(kx)l .....(1)
[positive denotes anticlockwise]
From the diagram, sinθ=xl2=2xl
θ is very small sinθθ.
where θ is expressed in radian.
Thus θ=2xl or x=θ2l.

From (1), we can write that
τ=k(θl2)×l=kθl22
But we know that τ=Iα
Moment of inertia of the rod about O I=ml212
Thus, we get
ml212α=kl22θ
α=6kmθ
Comparing with α=ω2θ we get,
ω2=6km
Time period of oscillations
T=2πω=π2m3k
Thus, option (c) is the correct answer.

flag
Suggest Corrections
thumbs-up
6
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon