# Showing SHM

1. Feb 28, 2014

### NATURE.M

1. The problem statement, all variables and given/known data

A cube of mass m is connected to two rubber bands of length L, each under tension T. The cube is displaced by a small distance y perpendicular to the length of the rubber bands. Assume the tension doesn't change. Show that the system exhibits SHM, and find its angular freqency ω.

3. The attempt at a solution

So basically from a FBD of cube, I have vertical forces: -2Tsinθ - mg = m$\frac{d^2y}{dt^2}$ and the horizontal components of tension from each band cancels. Now since the cube is displaced by a small distance y, I assume we can approximate sinθ ≈ θ. But then I'm not sure what to do?
I tried using sinθ = $\frac{y}{(y^2+L^2)^{1/2}}$, but then I get a complicated expression.
I know I need to obtain a -constant*y on the LHS. Any suggestions.

2. Feb 28, 2014

### NATURE.M

I think i got it, your assuming sinθ ≈ y/L, for small y.
One question though, do we have to assume gravity is negligible to get a sensible answer?

Last edited: Feb 28, 2014
3. Feb 28, 2014

### vela

Staff Emeritus
No, you don't. The differential equation becomes
$$y'' + \frac{2T}{mL}y + g=0.$$ Now consider a change of variables to $u = y+\frac{mg}{2T}L$. What's the differential equation in terms of $u$? What does $\frac{mg}{2T}L$ physically represent?

4. Feb 28, 2014

### NATURE.M

We haven't studied differential equations in much depth (since its a introductory physics course), so I didn't really catch the change of variables part. If you could explain further, I'd appreciate it.

5. Feb 28, 2014

### jackarms

I would just throw gravity out. The problem doesn't specifically mention it, so it might as well be on a frictionless tabletop or so.

6. Mar 1, 2014

### vela

Staff Emeritus
I'm saying rewrite the equation in terms of u instead of y. If you differentiate u twice, you get u''=y'', right? Just substitute in for y and y''.

It would be kind of hard to oscillate vertically on a flat tabletop.

7. Mar 1, 2014

### Staff: Mentor

When I read the problem, I also assumed (as jackarms did) that the oscillation was taking place in the horizontal, not the vertical. There is nothing in the problem statement that mentions the vertical. I pictured a horizontal frictionless table.

Chet

8. Mar 1, 2014

### Ronaldo95163

this is what I came out with...I also went on to show the frequency as well...

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9. Mar 1, 2014

### Staff: Mentor

This is definitely not what my answer would have been. I would have had T as a parameter in the frequency. I would have had:
$$f=\frac{1}{2π}\sqrt{\frac{2T}{mL}}$$
This is based on Vela's equation in post #3, with g removed.

Chet

10. Mar 1, 2014

### vela

Staff Emeritus
That's true. I inferred it from the use of $y$, but that's not really justified.

11. Mar 2, 2014

### Ronaldo95163

The angular frequency is actually represented by omega...f is the frequency of oscillation

12. Mar 2, 2014

### Staff: Mentor

Yes. ω=2πf

Chet

13. Mar 2, 2014

### Ronaldo95163

Yip so my proof of the angular frequency stops at the third to last line