SHM question involving a tunnel through earth

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The discussion revolves around a physics problem involving simple harmonic motion (SHM) of a travel pod dropped through a tunnel from Liverpool to New Zealand. The gravitational force acting on the pod is proportional to its distance from the Earth's center, leading to the equation z=R0cos(ωt). Participants seek to determine the pod's velocity at the Earth's center and the time it takes to reach the opposite side. There is confusion regarding the calculations, particularly concerning the velocity exceeding the speed of light, indicating a misunderstanding in the approach. The conversation highlights the importance of correctly applying SHM principles and equations to solve the problem effectively.
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Homework Statement


A tunnel is bored through the centre of the Earth from Liverpool to New Zealand and a travel pod is dropped into it. The gravitational force on the pod is proportional to its distance from the centre of the Earth and so it undergoes simple harmonic motion described by

z=R0cos\omegat

where

z=position relative to the centre of the Earth
R0=radius of the Earth=6*106m
\omega=1.2*10-3rads-1

1) What is the velocity of the pod when it reaches the centre of the Earth?

2) How long does the pod take to travel to the opposite side of the Earth?

Homework Equations


The one displayed above

The Attempt at a Solution


For question 1) since its at the centre of the Earth z=0 giving:
0=R0cos\omegat
cos\omegat=0
\omegat=1
t=1/\omega
and that's where I hit a dead end. I know my approach is completely wrong since the end result for velocity is faster than the speed of light however I don't know exactly how to remedy this situation. Any hints will be greatly appreciated as this is frustrating me how I can't answer a basic SHM question like this.
 
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I don't see how you calculated velocity at all.

What did you get for the time to reach the center of the Earth?

What's the period of this motion?

What is dz/dt as a function of t?
 
Doesn't matter, I worked it out now. Thanks for the reply anyway.
 
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