Find the Period of Oscillation for a Mass Falling Through a Tunnel Through the Earth

[NTesla]

Homework Statement

Kindly see the attached screenshot for Question. I'm trying to find out the time period of oscillation.

Relevant Equations

$a = - {omega}^2 . x$

I'm trying to solve it by 2 methods(screenshot below). However, I'm getting 2 different answers. I'm not able to find out the error in method 1. Method 2 gives the answer correctly. Kindly help.
P. S: If the pics are not clearly visible enough, kindly let me know. I'll try to either improve the pic or write using LaTeX.

Here's the question:

Here's my attempt:

 

[jbriggs444]

Homework Statement: Kindly see the attached screenshot for Question. I'm trying to find out the time period of oscillation.
Relevant Equations: $a = - {omega}^2 . x$

I'm trying to solve it by 2 methods(screenshot below). However, I'm getting 2 different answers. I'm not able to find out the error in method 1. Method 2 gives the answer correctly. Kindly help.
P. S: If the pics are not clearly visible enough, kindly let me know. I'll try to either improve the pic or write using LaTeX.

Here's the question: View attachment 347123

Here's my attempt:
View attachment 347124

Transcribing so that I can see what you are doing and comment effectively.

"Method 1...$$F = \frac{GMm}{x^2} \rightarrow a = \frac{F}{m} = -\frac{GM}{x^2}$$So far, so good. Mass $M$ will be a function of x. You are aware of this because the next line goes into a calculation of M.

"$$M=\frac{4}{3}\pi x^3 \rightarrow a = -\frac{G}{x^2} \times \frac{4}{3} \pi x^3$$ $$\rightarrow a = \frac{-4G\pi}{3} x \text{ m/s}^2$$Are you sure that the units come out right there? Does the density of the earth play any role?

If $x$ is a distance then you should not need to tack on units of meters.
If $x$ is a number of meters then you would need to tack on units of meters.

If $G$ is Newton's universal gravitational constant then you should not need to tack on units.
If $G$ is the numerical value of Newton's constant in some particular units then you would need to tack on units.

But you seem to be missing the forest for the trees. You have not been asked for numbers. You should not need to fret about the actual density of the Earth, its exact radius, the actual numeric value of Newton's gravitational constant or even about what system of units you will adopt.

Personally, I'd rather reason about this with words rather than equations. Proportionalities rather than equalities.

If you are halfway out from the Earth's center, what is your weight compared to your weight at the surface?

If you are fraction $x$ (between zero and one) of the way out from the center, what fraction of your surface weight would you be subject to?

 

[NTesla]

@jbriggs444
Thanks for your help. I understand that putting numerical values was not required in the question. However, I was trying to match the official answer. Official answer has numerical value.

 

[Nugatory]

Here's my attempt: [Image posted....]

Please please please learn to post using Latex. It only takes a moment, it makes life way easier for the unpaid volunteers who are generously offering their time to help you, and you will get much better answers.

@jbriggs444
I understand that putting numerical values was not required in the question. However, I was trying to match the official answer. Official answer has numerical value.

What is this "official answer" that you're talking about it? It's not in the problem statement in your original post.... And leaving stuff out of the problem statement almost guarantees that we'll be less helpful than we could be.

 

[kuruman]

Please please please learn to post using Latex. It only takes a moment, it makes life way easier for the unpaid volunteers who are generously offering their time to help you, and you will get much better answers.

Amen to that. To be fair to OP, it seems that there is an attempt to use LaTeX, albeit rudimentary, as seen in the "relevant equation"

Relevant Equations: $a = - {omega}^2 . x$

 

[NTesla]

What is this "official answer" that you're talking about it? It's not in the problem statement in your original post.... And leaving stuff out of the problem statement almost guarantees that we'll be less helpful than we could be.

The official answer was 5080 seconds, approximately 85 minutes.

I wasn't aware, that while posting questions, one also needs to post the official answer too. Also, answer is not part of the problem statement, so leaving it must NOT guarantee that anyone will be less helpful. How is helping someone contingent upon whether that person has posted the official answer too or not ? I would have agreed if you would have stated that leaving out the attempt to solve the question, almost guarantees that people will be less helpful or not helpful at all, but same could not be said about not posting the official answer alongwith the question.

I had mentioned about the official answer, only because I felt that the it needs some clarification as to why I was putting actual numerical values in the equation instead of working with the variable only.

 

[kuruman]

I had mentioned about the official answer, only because I felt that the it needs some clarification as to why I was putting actual numerical values in the equation instead of working with the variable only.

The problem is asking you to

There is no official answer. You just need to show that two times calculated in two different ways, give the same answer. Your calculations show that
Method 1: $~~T=\sqrt{\dfrac{3\pi}{G}}.$

Method 2: $~~T=2\pi\sqrt{\dfrac{R_E}{G}}.$

At this point you don't need to substitute numbers to see that the expressions are not equal. For the two times to be equal, it must be true that $$(2\pi)^2 R_E=3\pi\implies R_E=\frac{3}{4\pi}$$which is clearly not the case. So one or both of your calculations are incorrect. You know that Method 1 is flawed because the calculated time is dimensionally incorrect since $\sqrt{1/G}$ does not have dimensions of time. To see if Method 2 gives the correct result, you substitute numbers and indeed you get the "official" numerical answer which is not part of the problem description. Doing that is OK. That answer was provided to help you. So where does that leave you?

You need to fix Method 1 and obtain the correct algebraic expression which now you know is $~~T=2\pi\sqrt{\dfrac{R_E}{G}}.$ Start by fixing $M$, what you call "the mass of earth within radius $x$." You write $M=\frac{4}{3}\pi x^3$. On the left side of the equation I see a mass, but on the right side I see a volume.

 

[NTesla]

The problem is asking you to
View attachment 347407
There is no official answer. You just need to show that two times calculated in two different ways, give the same answer. Your calculations show that
Method 1: $~~T=\sqrt{\dfrac{3\pi}{G}}.$

Method 2: $~~T=2\pi\sqrt{\dfrac{R_E}{G}}.$

At this point you don't need to substitute numbers to see that the expressions are not equal. For the two times to be equal, it must be true that $$(2\pi)^2 R_E=3\pi\implies R_E=\frac{3}{4\pi}$$which is clearly not the case. So one or both of your calculations are incorrect. You know that Method 1 is flawed because the calculated time is dimensionally incorrect since $\sqrt{1/G}$ does not have dimensions of time. To see if Method 2 gives the correct result, you substitute numbers and indeed you get the "official" numerical answer which is not part of the problem description. Doing that is OK. That answer was provided to help you. So where does that leave you?

You need to fix Method 1 and obtain the correct algebraic expression which now you know is $~~T=2\pi\sqrt{\dfrac{R_E}{G}}.$ Start by fixing $M$, what you call "the mass of earth within radius $x$." You write $M=\frac{4}{3}\pi x^3$. On the left side of the equation I see a mass, but on the right side I see a volume.

Thank you for your detailed response. It's sincerely very much appreciated.

However, I had figured where I was going wrong after reading jbriggs444's answer in post #2.

And there is official answer with numerical value. Here's the screenshot of the official answer.

It was interesting to see 85 minutes as the answer, as it reminded me that the International space station(ISS) revolves around earth in approximately 90 minutes. So, it's not irrelevant to get to the numerical value of the answer. It is a kind of memory peg, that if there was a tunnel(hypothetical offcourse) through the earth, then it's time period would be that of ISS(approximately). If the mass in the tunnel was released just when the ISS was flying over it, the people in the ISS would see the mass returning to the point of release, everytime they returned to that point in space.