How Do You Derive Coupled Pendulum and Mass on Spring Equations of Motion?

[mr.hood]

Homework Statement

A mass M₁ sits on a level frictionless surface attached to a spring of stiffness k. Mass M₂ is supported from M₁ by a string of length $$l$$. The horiztonal displacement from the equilibrium position of M₁ is x₁; the horizontal displacement of the pendulum from the wall where the spring is attached is x₂; and the vertical displacement of the pendulum is y₂. Find the exact equations of motion for x₁, x₂, and y₂. Then, show that the equations can be reduced, using the small angle approximation (sin(x) ~ tan(x) ~ x) and the fact that
$$\ddot{y}_{2}<<g$$

to:

$$M_{1} \ddot{x}_{1} = -k x_{1} + M_{2} \frac{g}{l} \left(x_{2} - x_{1}\right)$$
$$M_{2} \ddot{x}_{2} = - M_{2} \frac{g}{l} \left(x_{2} - x_{1}\right)$$

Homework Equations

The Attempt at a Solution

So far, I've attempted to derive the exact equations for motion:

$$M_{1} \ddot{x}_{1} = -k x_{1} + M_{2} g \left(sin\theta cos\theta \right)$$
$$M_{2} \ddot{x}_{2} = -M_{2} g \left(sin\theta cos\theta \right)$$
$$M_{2} \ddot{y}_{2} = -M_{2} g \left(sin^{2} \theta \right)$$

where theta is the angular displacement of the pendulum from the equilibrium position. In the first equation, the force terms on the right come from the spring and the horizontal component of the tension in the string connecting the two masses (which is the component of the weight of the pendulum pointed parallel to the string). In the second and third equations, the force terms are just the horizontal and vertical components of the component of the pendulum's weight pointed in the direction of its motion. I find it would be much easier to just derive the equations of motion in terms of the angle theta, and then use the small angle approximation, but I have to do it from the equations I just gave. I'm just having trouble making the leap to the simplified equations. Any ideas?

 

[Jhero]

Hi there,

Thank you for your post and for sharing your attempt at the solution so far. It looks like you are on the right track in deriving the exact equations of motion for the system.

To simplify the equations using the small angle approximation, we need to assume that the angle theta is very small (i.e. sin(theta) ~ theta). This allows us to rewrite the equations as:

M_{1} \ddot{x}_{1} = -k x_{1} + M_{2} g \left(\theta cos\theta \right)
M_{2} \ddot{x}_{2} = -M_{2} g \left(\theta cos\theta \right)
M_{2} \ddot{y}_{2} = -M_{2} g \left(\theta^{2} \right)

Next, we can use the fact that cos(theta) ~ 1 and theta^2 << g to further simplify the equations to:

M_{1} \ddot{x}_{1} = -k x_{1} + M_{2} g \theta
M_{2} \ddot{x}_{2} = -M_{2} g \theta
M_{2} \ddot{y}_{2} = 0

Finally, we can substitute in the small angle approximation (sin(theta) ~ theta) and the fact that \ddot{y}_{2} << g to get the simplified equations of motion:

M_{1} \ddot{x}_{1} = -k x_{1} + M_{2} \frac{g}{l} \left(x_{2} - x_{1}\right)
M_{2} \ddot{x}_{2} = - M_{2} \frac{g}{l} \left(x_{2} - x_{1}\right)

I hope this helps! Let me know if you have any further questions or if you need any clarification.