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

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R2) from the earth's centre, where R is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:

A
2πRg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
12πgR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2πRg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
g2πR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2πRg
cos θ=xd

If displaced from equilibrium position,

Frestoring=(GMmdR3)cosθ

Frestoring=GMmdR3.xd=GMmxR3

a=GMxR3

we know that the time period given by,

T=2πxa

T=2π xGMxR3

T=2πRg (g=GMR2)

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Physical Pendulum
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon