Solving SHM of Cube Connected By Rubber Bands

[NATURE.M]

Homework Statement

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.

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.

 

[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?

 

[vela]

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?

 

[NATURE.M]

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?

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.

 

[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.

 

[vela]

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.

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''.

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.

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

 

[Chestermiller]

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

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

 

[Ronaldo95163]

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

 

[Chestermiller]

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

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

 

[vela]

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

 

[Ronaldo95163]

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

 

[Chestermiller]

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

Yes. ω=2πf

Chet

 

[Ronaldo95163]

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