Is the Time to Reach Earth's Center Correct?

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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)
IMG_20200727_173222.jpg


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?
 
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tanaygupta2000 said:
so that x'' = (w2)x
Watch the signs.
tanaygupta2000 said:
but the question is asking only for the half journey
What would one complete cycle involve?
 
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.)
 
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