# Simple Harmonic Motion

1. Apr 25, 2007

### kingwinner

1) A mass is connected to 2 rubber bands of length L as shown. Each rubber band has a constant tension T. You can neglecct the force of gravity for this problem. The mass is displaced horizontally by a very small distance x and then released (so x<<L always). The mass will exhibit simple harmonic motion. Express your answer in terms of the variables given in the quesiton. (the picture is not to scale, x is actually much smaller). Find a simple expression for the small angle theta in terms of the displacement x of the mass using the fact that x<<L.

I know how to solve SHM problems for a spring, but I am so lost with this problem...how should I start? Can someone kindly give me some help/hints? Thanks a lot!

2. Apr 25, 2007

### RoseCrye

Only thing I can think of to do with this problem is to state it as a tangent... that would give you theta.

Can you do refraction problems?

3. Apr 25, 2007

### mjsd

first: draw a force diagram
second: relate all variables using equations that you derived from the diagram
third: solve for theta in the small angle approximation: which usually means
$$\sin \theta\approx \theta$$ and $$\cos\theta \approx 1$$

4. Apr 25, 2007

### kingwinner

I can't do refraction problems.

5. Apr 25, 2007

### kingwinner

When we move the particle out a little bit from the equilibrium (middle), the tensions in the vertical component cancels, and the horizontal component adds together...but I can I relate it to "x"? and how can I find an expression for theta?

sin theta = x/L

How can I use the fact that x<<L? (binominal approximation? small angle approximation?)

I am still kind of lost...

6. Apr 25, 2007

### mezarashi

Are you expected to solve second order differential equations for this course?

7. Apr 26, 2007

### kingwinner

No, I haven't learnt how to solve differential equation

8. Apr 26, 2007

### nrqed

That is right. Notice that when x is positive, the net force is to the left so the net force will be $-2 T sin \theta$
You know what? With the drawing shown, you would not need the approximation!

usually, this problem is given with L being measured along the unstretched string. In that case, sin of theta would be given by $sin \theta = \frac{x}{\sqrt{x^2 +L^2}}$ You see why the approximation x<<L would be required to simplify things . But with the L shown as it is in the drawing, there is no need to use the approximation.

The final step is to compare your equation to the equation for a mass attached to a spring and you can see what expression plays the role of the spring constant "k" (you can then find the angular frequency or the period of oscillation)

9. Apr 26, 2007

### e(ho0n3

Good. So you know that there is a restoring force equal twice the horizontal component of the tension. What you need to do now is show how how this restoring force is related to the horizontal distance moved from equilibrium.

For example, let F(a) be the restoring force when the mass is displaced horizontally by a distance a. Then, as you have stated, F(x) equals twice the horizontal component of the tension on one of the rubber bands. Is F(a) proportional to a? What happens when you treat the rubber bands like springs?

I would use the small angle approximation (given by mjsd).

10. Apr 26, 2007

### kingwinner

For small theta,
sin(theta) = theta = x/L (we need to eliminate L)

Now if I can I find a way of writing L in terms of x only, then I am done, but how can I do so?

11. Apr 26, 2007

### nrqed

You can't. Your final expression will contain L. Why do you want to get rid of L?