# Simple harmonic motion problem

1. Apr 4, 2007

### fruitl00p

1. The problem statement, all variables and given/known data
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?
2. Relevant equations

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

3. 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?

2. Apr 4, 2007

### Andy_ToK

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)

3. Apr 5, 2007

### SGT

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.

4. Apr 5, 2007

### Mentz114

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.

Last edited: Apr 5, 2007
5. Apr 5, 2007

### fruitl00p

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

6. Apr 5, 2007

### fruitl00p

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

7. Apr 5, 2007

### Mentz114

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).

Last edited: Apr 5, 2007
8. Apr 5, 2007

### e(ho0n3

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

9. Apr 5, 2007

### e(ho0n3

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.

10. Apr 5, 2007

### Mentz114

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.

Last edited: Apr 5, 2007
11. Apr 5, 2007

### fruitl00p

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?

12. Apr 5, 2007

### Mentz114

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

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

13. Apr 5, 2007

### fruitl00p

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?

14. Apr 5, 2007

### fruitl00p

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

15. Apr 5, 2007

### Mentz114

16. Apr 5, 2007

### fruitl00p

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?

17. Apr 5, 2007

### fruitl00p

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.

18. Apr 5, 2007

### Mentz114

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

19. Apr 5, 2007

### fruitl00p

my calculator should be in radians, correct?

20. Apr 5, 2007

### fruitl00p

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

Yet, I still cannot get part two correct