# Simple harmonic motion of small planet

## Homework Statement

The residents of a small planet have bored a hole straight through its center as part of a communications system. The hole has been filled with a tube and the air has been pumped out of the tube to virtually eliminate friction. Messages are passed back and forth by dropping packets through the tube. The planet has a density of 4120kg/m^3, and it has a radius of R= 5.67E+6 m. What is the speed of the message packet as it passes a point a distance of .420R from the center of the planet?
How long does it take for a message to pass from one side of the planet to the other?

## Homework Equations

1.density=mass/volume= mass/[(4/3)pi*R^3]
2. x=Acos(wt)
3. v=-wAsin(wt)

## The Attempt at a Solution

I first found the mass in equation one. Then I thought .420R was x, the radius is A and, that is as far as I got with that.

Am I wrong so far?

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Since it's simple harmonic, you can find (acceleration) a=w^2*x first. (express the gravitational force exerting on the packet as a linear function of r(the distance between the packet and the center of the planet)

SGT
Since it's simple harmonic, you can find (acceleration) a=w^2*x first. (express the gravitational force exerting on the packet as a linear function of r(the distance between the packet and the center of the planet)
Don´t forget that the mass of the planet to be considered in the calculation of the gravitational force is the contained in the sphere of radius r.

The only mass exerting a nett force on the packet is that contained inside the current radius because the shell of matter outside the radius has no effect ( as pointed out above). So the force at point x is

$$F(x) = -\frac{4}{3}\pi Gm\rho x^3/x^2$$

which gives

$$F(x) = -Kx$$

which is the equation for a simple harmonic oscillator.

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Mentz114, so m in that equation is the mass of the packet? Because in this problem we are not given the mass of the packet, and that has been confusing me. (well a lot of this problem period is confusing me). I figured that since the mass of the packet wasn't mentioned then it somehow gets canceled out in the problem.

Andy_ToK, I don't understand how to find the acceleration. Using the equation Mentz114 suggests, how do I find the acceleration?

I tried at first to set ma=GMm/R^3 and then the m of packet cancels out.

Our class just started doing simple harmonic motion yesterday so I am still trying to understand the concepts. This problem is just so confusing.

So Mentz114, am I to assume that x in F= -kx, is .420R in my problem? and if so then the k is -\frac{4}{3}\pi m

ok the last part I meant k = -(4/3) pi*m*p

Hi Fruit,

I made an error and left out a 'G' in the force formula,

$$F(x) = -\frac{4}{3}\pi Gm\rho x^3/x^2$$

You'll find from your SHO studies that the frequency of oscillation depends only on K and m so if you have K you can work out the time part.

By the way, acceleration is force/mass ( Newton).

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Wouldn't it be easier to solve this problem using energy methods once k is determined?

$$F(x) = -\frac{4}{3}\pi Gm\rho x^3/x^2$$
Shouldn't that be

$$F(x) = -\frac{4}{3}\pi Gm\rho R^3/x^2$$

although, if it were, the motion of the packet would not be simple harmonic.

Only the matter inside the current position ( ie with r < x) exerts a nett force. The remaining matter forms a hollow shell which, as you know exerts no field inside.

See post #3 and #4.

Using energy we can get the maximum PE of the oscillator and equate to the maximum grav potential, so

$$KR^2 = \frac{GMm}{R}$$

so

$$K = \frac{GMm}{R^3}$$

and

$$K = \frac{4}{3}\pi Gm\rho$$

which is the same as above.

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oh, ok. so am I right to assume that when I set up the equation for the period, 2pi * sqrt m/k, that the m's will cancel?

Yes, you got it. I think this might be the only gravitational field that can give harmonic oscillations.
To solve the velocity part of the question you'll need the solution to

$$\frac{dx}{dt} = f(K, x)$$

I keep getting the period incorrect
I did 2*pi *sqrt m/GMm/R^3, I canceled out the m's
leaving 2*pi *sqrt R^3/G*M = 2pi sqrt (5.67E6)^3/[(6.67E-11)*(3.145E24)]
PERIOD = 5856 s

Plus, a friend just told me to try a= GM/R^2, then find angular speed
with angular speed = sqrt a/R.
Since the period is 2pi/angular speed as well, I used the answer I got and got the same thing!

For some reason I still cannot get the first part of the problem.

Please tell me if my rationale on this portion is wrong: x=Acos(wt)
so the x would be the .420R, A would be simply R and the angular speed is w.
when I tried the k Mentz114 suggested I got the same answer that I got when I tried w=sqrt a/R
so I think I'm doing it correctly but when I tried to find t, I think I solved it incorrectly....I'm not the greatest at math
I did x/A*cosw which became x/cos w since the R's cancel. I ended up getting a value of t= .4200s However, when I put this value in the velocity equation, v= -wAsin(wt), I got a small negative velocity that was wrong.

and to find the mass of planet, it is density times volume and the forumla for the volume is 4/3 pi R^3...I just don't understand what I am doing wrong?

Hmm... I understand that dx/dt is the dirst derivative of the position equation, but I do not understand what f(K,x) means

Mentz114, I guess I didn't do a good job at explaining what I thought you meant. Isn't the velocity equation the first derivative of the position equation? That was all I meant.

I am looking at the equations, but I don't know what t is, and I don't know how to solve for t

Could I actually use energy conservation in this problem?

I just tried to do 1/2 mv^2= 1/2 kA^2
v^2 = (GM/R^3)*A^2
v= sqrt (GM/R^3) *A
and got v=6.083 *10^3 m/s and that was wrong.

What is the speed of the message packet as it passes a point a distance of .420R from the center of the planet?
Use the position equation to solve for t, then use t in the velocity equation.

0.42R= Rsin(wt)
sin(wt) = 0.42
wt = sin^-1(0.42)
t = sin^-1(0.42)/w

w = 1/T

my calculator should be in radians, correct?

Mentz114, thank you, I was able to get the correct answer for part one.

Yet, I still cannot get part two correct

Could I actually use energy conservation in this problem?
Yes you can. Once you know k, you can find the velocity at some distance r as follows:

$$mv^2 + \frac{1}{2}kr^2 = \frac{1}{2}kR^2$$

for part two, I thought that the question was asking for the period, but no, just how long it takes to get from one side to the other....and I'm stuck.
I'm trying to use the position equation to get it but I'm not getting very far

Doc Al
Mentor
for part two, I thought that the question was asking for the period, but no, just how long it takes to get from one side to the other....and I'm stuck.
But how does the needed time relate to the period?

Doc Al, I realized that I was making that section of the problem complicated! I figured out how to do it but just realizing that the time it takes from one side to the other is just 1/2 of the period.