# Simple Harmonic Motion

Discussion in 'Physics & Math' started by kingwinner, Apr 25, 2007.

1. ### kingwinnerRegistered Senior Member

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796
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.

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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!

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3. ### temurman of no wordsRegistered Senior Member

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The elastic force is proportional to the change in the rubber's length. You can calculate this change depending on x and then take into account that x is small. This is usually done by a truncated Taylor expansion. Don't forget to project the force onto the horizontal line and multiply by two to account the second string.

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5. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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temur, note that the change in the length of the rubber band is second order in x and hence cannot be responsible for the linear restoring force. This is hinted at in the question statement where we are instructed to consider the rubber bands with a constant tension. The force comes just from the projection.

kingwinner, the force due to each rubber band is constant in magnitude, but it can change direction. When the particle is not displaced at all, each rubber band pulls the particle up or down respectively and the forces cancel. What happens to the direction of each force when we move the particle out a little bit? Quantify this effect in terms of x and you will have your restoring force.

Hope this helps.

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7. ### kingwinnerRegistered Senior Member

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796
When we move the particle out a little bit, 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...

Thanks for your help!

8. ### temurman of no wordsRegistered Senior Member

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Physics Monkey, thanks, I did not see that coming.

kingwinner, you can just use sin(theta) = x/L, since the projection of T on the horizontal axis is T*sin(theta). For small angle, theta = sin(theta).

9. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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kingwinner, just calculate the component of the force in the horizontal direction. The fact that x << L is what enables you to neglect the change in the length of the rubber band.

The diagram is a bit confusing. If I call L0 the length of the rubber band when x = 0, then simple geometry will tell you that L = srqt(x^2 + L0^2). The fact that x << L means that x << L0 and so L0 is approximately equal to L. The correction is of order x^2 and hence is very small. Now hopefully its clear that you need to use the small x approximation to make L independent of x.

10. ### leopoldValued Senior Member

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17,455
doesn't this suggest a trick question?

11. ### kingwinnerRegistered Senior Member

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796
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?

12. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
I don't think you can get rid of L like that.
But you might be able to get rid of it using m (mass of the object) and T (tension) instead?

13. ### kevinalmRegistered Senior Member

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993
Since theta is very small L and T can be taken as constants. Then using the x = sin x small angle approximation:

F/x = - T/L
F = x *(- T/L)

You don't want to eliminate T and L, T/L is your spring constant so to speak.