# Homework Help: Oscillation question 2 (SHM)

1. Jan 16, 2014

### xtrubambinoxpr

I apologize ahead of time for all of these post about oscillation. I am trying to learn this stuff on my own.

I answered "a" because the x^2 function. But I don't know the second part.. Would it also be 3x^2 because it is proportional to the x(t) function?

2. Jan 16, 2014

### nasu

What type of force acts in a SHO? What is the dependence on displacement?

3. Jan 16, 2014

### xtrubambinoxpr

depends on the force being acted on it?

4. Jan 16, 2014

### nasu

What depends on the force????? I don't understand this.

Imagine a simple harmonic oscillator. You displace it from equilibrium by a distance x.
What force acts to bring it back to equilibrium? What is the dependence of this force of x?
You can think about a specific system, like a ball attached to a spring. In this case the force is the elastic force isn't it?

5. Jan 16, 2014

### xtrubambinoxpr

The gravity would be pulling it down and the force would be opposite.
Or with a block the spring is applying the force if it's on a table both ways. Acceleration will be applied opposite of the motion. So the force is given by the spring

6. Jan 16, 2014

### Staff: Mentor

Good. If the spring stretches by x, what is the force that the spring exerts on the mass? From Newton's second law, what is the acceleration (as a function of the spring constant, the mass of the mass, and the displacement from equilibrium x)?

7. Jan 16, 2014

### xtrubambinoxpr

You lost me. Only using the spring formulas or the oscillation ones? I don't see the connection. If F=kx. Then kx/m=a

8. Jan 16, 2014

### Staff: Mentor

Actually, it looks like you do see that connection. You just didn't realize it. These are the equations I was asking about. Actually, the equation for the acceleration should have a minus sign, because the spring is a restoring force constantly trying to bring the mass back to the equilibrium position; but the mass always overshoots the equilibrium position. Back to the equation for the acceleration: a = -kx/m. Look at your problem statement now, and see which relationship between the acceleration and the displacement agrees with this equation.

Now, back to the relationship between this and SHM. The acceleration is the second derivative of the displacement with respect to time, so we have:
$$a=\frac{d^2x}{dt^2}=-\frac{k}{m}x$$
The complimentary solution to this differential equation is:

x=A sin(ωt) + Bcos(ωt)

where $ω=\sqrt{\frac{k}{m}}$

The constants A and B in the equation depend on the initial displacement and the initial velocity of the mass.

Do you see now how this ties in with SHM?

Chet

9. Jan 17, 2014

### xtrubambinoxpr

So, if I understand this completely, I was wrong because there is no x^2 in regards to the acceleration. it would be the -4x option because it seems to be closely related to what we just proved. I think I was thinking of ω$^{2}$

10. Jan 17, 2014

### sophiecentaur

There's a very important minus sign missing in your Force formula.

Have you been set this question without any prior information about SHM and do you have no text to work from?

11. Jan 17, 2014

### xtrubambinoxpr

I have the text and I now remember the (-) sign in the equation. Completely my mistake. I have 2 books I am trying to learn from so forgive me if I forget something.

12. Jan 17, 2014

### sophiecentaur

Your book (or Wiki) will have it all in there - you may just need to read it more slowly.

What is the relationship (always) between force and acceleration and how can you relate that to your Force equation (with the right sign included). Now look at those alternatives in the question and find one that has the same form as this.

PS Using angular frequency ω, is always annoying and perplexing to start with but it is very useful to get used to it because, once you start to get advanced, if you continue to use frequency f, the constant 2∏ keeps cropping up, multiple times and just gets in the way.

13. Jan 17, 2014

### xtrubambinoxpr

based on what you're saying now I think it is either 5x or -4x. the force is directly proportional to the acceleration. I think I am missing the final key point

14. Jan 17, 2014

### xtrubambinoxpr

By the way if anyone were to want to skype or google hangout on this topic I would be cool with that too. Just trying to learn.

15. Jan 17, 2014

### sophiecentaur

Why was I so picky about the sign? It tells you the direction that the force is acting in and is essential in the mathematical description of the motion. Which direction is the force? To reduce or increase the magnitude of displacement?

16. Jan 17, 2014

### xtrubambinoxpr

it is negative because it is the restoring force to get it back to its initial position

17. Jan 17, 2014

### sophiecentaur

So which of the alternatives is showing that?

18. Jan 17, 2014

### xtrubambinoxpr

-4x is the only one abiding by that principle

19. Jan 17, 2014

### sophiecentaur

So, how does that look? A reasonable answer?

20. Jan 17, 2014

### xtrubambinoxpr

being that it needs to have the equation form to fit, yes I say it is reasonable. a(t) = -ω$^{2}$x(t).. that being said -4x is the only option that seems to fit.

21. Jan 17, 2014

### sophiecentaur

That is a fresh addition. Can you explain where it came from? (I am trying to do this one step at a time.)

22. Jan 17, 2014

### xtrubambinoxpr

the position function for a particle in SHM is x(t) = Xmcos(ωt+ø)

differentiating it you get velocity. and again you get a(t) = -ω$^{2}$XmCos(ωt+ø) .... = a(t) = -ω$^{2}$x(t)

23. Jan 17, 2014

### sophiecentaur

Right. So that tells you that ω2 equal to what? And hence ω= what?

24. Jan 17, 2014

### Staff: Mentor

Correct. Nice job of figuring it out.

Chet

25. Jan 17, 2014

### xtrubambinoxpr

would it be Sqrt(a(t)/x(t))

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