Home
Class 11
PHYSICS
if a boby is released into a tunal dug ...

if a boby is released into a tunal dug across the diameter of earth, it executes simple harmonic motoin with time period

A

`T=2pisqrt((R_(e))/(g))`

B

`T=2pisqrt((2R_(e))/(g))`

C

`T=2pisqrt((R_(e))/(2g))`

D

`T=2` seconds

Text Solution

Verified by Experts

The correct Answer is:
A

Length of pendulum=`R_(e)`
`therefore T=2pisqrt((R_(e))/(g))`.
Promotional Banner

Similar Questions

Explore conceptually related problems

If a body is executing simple harmonic motion, then

The displacement of a particle in simple harmonic motion in one time period is

The velocity of a particle executing simple harmonic motion is

While a particle executes linear simple harmonic motion

The total energy of a particle executing simple harmonic motion is

The distance moved by a particle in simple harmonic motion in one time period is

The phase of a particle executing simple harmonic motion is pi/2 when it has

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R//2) 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 solid sphere of mass M and Radius R is attached to a massless spring of force constant k as shown in the figure. The cylinder is pulled towards right slightly and released so that it executes simple harmonic motion of time period. If the cylinder rolls on the horizontal surface without slipping, calculate its time period of oscillation.

Imagine a tunnel dug along a diameter of the earth. Show that a particle dropped from one end of the tunnel executes simple harmonic motion. What is the time period of this motion? Assume the earth to be a sphere of uniform mass density (equal to its known average density=5520 kg m^(-3) .)G= 6.67xx10^(-11)Nm^(2)kg^(-2) .Neglect all damping forces.