Time taken to reach the centre

[tanaygupta2000]

Homework Statement

A tunnel is dug from the surface of the earth through the center and opens at the other end. A ball is dropped from one end of the tunnel. The acceleration due to gravity on the earth’s surface is g and the radius of the earth is R. Assuming that the earth has a constant density, what is the time taken by the ball to reach the center of the earth?

Options:
(a) π√(R/g)
(b) 2π√(R/g)
(c) (π/2)√(R/g)
(d) (π/4)√(R/g)

Relevant Equations

Acceleration due to gravity inside the earth, g' = g(1 - d/R)

The value of acceleration due to gravity at a depth 'd' inside the Earth is given by-
g' = g(1 - d/R)
which can also be written as
g' = g(x/R) from the diagram

so that x'' = (w²)x
where w² = g/R is the angular frequency

Hence the time period T is given by
T = 2π sqrt(R/g)

but the question is asking only for the half journey
so the answer should be
T = π sqrt(R/g)
Is this correct?

 

[haruspex]

so that x'' = (w2)x

Watch the signs.

but the question is asking only for the half journey

What would one complete cycle involve?

 

[tanaygupta2000]

 

[tanaygupta2000]

View attachment 266934

Is this approach correct ?

 

[PeroK]

Is this approach correct ?

Looks good.

 

[TSny]

As @haruspex pointed out, you have a sign error. The differential equation $\ddot x = \omega^2 x$ does not have $\sin \omega t$ and $\cos \omega t$ as solutions, as you can check.

However, the rest of your work looks good. (Regarding your first approach in post #1, traveling from the surface to the center of the Earth is 1/4 of a cycle.)