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

In summary: You can express everything in terms of L and then substitute L with a numerical value later. Or you can use the small angle approximation to eliminate theta and get an expression in terms of x only.In summary, the problem involves a mass connected to two rubber bands of length L, with a constant tension T and displacement x. The mass exhibits simple harmonic motion and the small angle theta can be expressed as x/L with the use of the small angle approximation. The restoring force is equal to twice the horizontal component of the tension, and for small theta, sin(theta) = theta = x/L. L cannot be eliminated from the final expression, but can be substituted with a numerical value later.
  • #1
kingwinner
1,270
0
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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  • #2
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
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
[tex]\sin \theta\approx \theta[/tex] and [tex]\cos\theta \approx 1[/tex]
 
  • #4
I can't do refraction problems.
 
  • #5
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!
 
  • #6
Are you expected to solve second order differential equations for this course?
 
  • #7
mezarashi said:
Are you expected to solve second order differential equations for this course?

No, I haven't learned how to solve differential equation
 
  • #8
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 [itex] -2 T sin \theta [/itex]
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 [itex] sin \theta = \frac{x}{\sqrt{x^2 +L^2}} [/itex] 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
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?
 

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and is directed towards the equilibrium point. In other words, it is a repetitive motion where the force acting on an object is directly proportional to its displacement from the equilibrium position.

What are some examples of Simple Harmonic Motion?

Some examples of Simple Harmonic Motion include the motion of a pendulum, the motion of a mass attached to a spring, and the motion of a mass on a horizontal surface with friction. In all these cases, the restoring force is directly proportional to the displacement from equilibrium, resulting in SHM.

What is the formula for Simple Harmonic Motion?

The formula for Simple Harmonic Motion is F = -kx, where F is the restoring force, k is the spring constant or stiffness of the system, and x is the displacement from equilibrium. This formula applies to all systems that exhibit SHM.

What is the relationship between Simple Harmonic Motion and energy?

In Simple Harmonic Motion, there is a constant exchange of potential and kinetic energy. When the object is at the equilibrium point, it has maximum potential energy and zero kinetic energy. As it moves away from equilibrium, potential energy decreases and kinetic energy increases until it reaches the turning point, where potential energy is zero and kinetic energy is maximum. The process then repeats in the opposite direction.

How does Simple Harmonic Motion differ from other types of motion?

Simple Harmonic Motion differs from other types of motion in that it is a special type of periodic motion where the restoring force is directly proportional to displacement. Other types of motion, such as circular motion or linear motion, may have different relationships between force and displacement.

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