How Do You Calculate Theta in Simple Harmonic Motion with Rubber Bands?

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The discussion focuses on calculating the angle theta in simple harmonic motion involving a mass connected to two rubber bands. Participants suggest using the small angle approximation, where sin(theta) can be approximated as theta, leading to the relationship sin(theta) = x/L. The conversation emphasizes the importance of understanding the restoring force, which is proportional to the horizontal displacement x, and how it relates to the tension in the rubber bands. Ultimately, the final expression for theta will include L, and participants clarify that eliminating L is unnecessary for solving the problem. The discussion highlights the connection between the rubber bands' behavior and spring dynamics in simple harmonic motion.
kingwinner
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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.
http://www.geocities.com/asdfasdf23135/phyexam1.JPG

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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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?
 
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
 
I can't do refraction problems.
 
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...

Thanks for your help!
 
Are you expected to solve second order differential equations for this course?
 
mezarashi said:
Are you expected to solve second order differential equations for this course?

No, I haven't learned how to solve differential equation
 
kingwinner said:
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
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
How can I use the fact that x<<L? (binominal approximation? small angle approximation?)

I am still kind of lost...

Thanks for your help!
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)
 
kingwinner said:
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"?

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?

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 would use the small angle approximation (given by mjsd).
 
  • #10
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
kingwinner said:
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?

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