# Period of Motion from Equations of Motion

I would like to preface this by saying that I solved the homework problem, but my professor gave me an added challenge of finding the period of the motion described in this problem.

1. Homework Statement

The pendulum bob of mass m shown in the figure below is suspended by an in-extensible string from point p. This point is free to move along a horizontal line under the action of the springs, each having an elastic constant k. Assume that point p is displaced a distance x to the right and then released; determine the equations of motion of the system.

I don't have an exact diagram so here's a replica with all necessary labels:

2. The Equations of Motion
After setting up the Lagrangian and finding the equations of motion (which have been checked by my professor), we get
$$m\ddot{x} + mr\ddot{\theta}\cos{\theta} - mr{\dot{\theta}}^2\sin{\theta} + 2kx = 0$$
$$\ddot{x} + r\ddot{\theta} + g\sin{\theta} = 0$$

## The Attempt at a Solution

First of all, we are assuming theta is very small. So far we have used an integral involving the mechanical and potential energy of a system to find the time it takes to travel from one position to another. The integral is essentially ##dt## written in terms of ##dx, E, V(x).## The problem is, I don't know what the maximum value of theta in this oscillation is. My professor thinks there is a way to make the period pop out of the equations under the small-angle approximations. After these approximations, I get

$$m\ddot{x} + mr\ddot{\theta} - mr{\dot{\theta}}^2\theta + 2kx = 0$$
$$\ddot{x} + r\ddot{\theta} + g\theta = 0$$

What immediately stood out to me was that both the ##x, \theta## accelerations could be canceled out in one step, giving ##2kx = m\theta(g + r{\dot{\theta}}^2)##. As I am writing this, I realize that separating the variables could be used here, and even better, that x can be written as an oscillator (duh).
So now I'm here:
$$\dot{\theta} = \sqrt{\frac{f(t)}{\theta} + C}$$ for some function of time ##f(t)## and some constant ##C##.
I don't know how to proceed, unless there is a way to solve this after finding f. Is there a more general way to do this given the equations of motion?

Thanks.

#### Attachments

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## Answers and Replies

TSny
Homework Helper
Gold Member
After setting up the Lagrangian and finding the equations of motion (which have been checked by my professor), we get
$$m\ddot{x} + mr\ddot{\theta}\cos{\theta} - mr{\dot{\theta}}^2\sin{\theta} + 2kx = 0$$
$$\ddot{x} + r\ddot{\theta} + g\sin{\theta} = 0$$
Before making the small-angle approximation, I believe that the first term of the second equation should include a factor of ##\cos \theta##.

After these approximations, I get

$$m\ddot{x} + mr\ddot{\theta} - mr{\dot{\theta}}^2\theta + 2kx = 0$$
$$\ddot{x} + r\ddot{\theta} + g\theta = 0$$
Looks good. But there's a term in one of these equations that can be neglected in the small-angle approximation. Throwing out this term will make life much easier.

Abourque72
Thanks for the response. I looked back and sure enough I was missing that cosine. Anyways, your suggestion makes me think that the term ##{\dot{\theta}}^2 \theta## should certainly be negligible since, if ##\theta## is small, ##\dot{\theta}## ought to be pretty small too. So this term is a "cubic" term while everything else is "linear".

Under this assumption, ##\theta## is proportional to ##x##, so the period of the angle and the movement of the pendulum are equal. That seems to make sense, but I am not sure if it is correct.

TSny
Homework Helper
Gold Member
Thanks for the response. I looked back and sure enough I was missing that cosine. Anyways, your suggestion makes me think that the term ##{\dot{\theta}}^2 \theta## should certainly be negligible since, if ##\theta## is small, ##\dot{\theta}## ought to be pretty small too. So this term is a "cubic" term while everything else is "linear".
Yes

Under this assumption, ##\theta## is proportional to ##x##
Yes. So, what do you get for the differential equation for ##x## as a function of time?

Abourque72
I'm assuming you can treat the mass m of the pendulum as a mass at point p, and since Hooke's Law makes the forces on the point in the same direction, its just a simple harmonic oscillator with spring constant 2k and mass m: $$\ddot{x} = \frac{-2kx}{m}$$

Hence, the period of motion is $$\pi \sqrt{\frac{2m}{k}}$$

TSny
Homework Helper
Gold Member
I'm assuming you can treat the mass m of the pendulum as a mass at point p, and since Hooke's Law makes the forces on the point in the same direction, its just a simple harmonic oscillator with spring constant 2k and mass m: $$\ddot{x} = \frac{-2kx}{m}$$

Hence, the period of motion is $$\pi \sqrt{\frac{2m}{k}}$$
No. You will need to use your differential equations from the Lagrangian.

Abourque72
No. You will need to use your differential equations from the Lagrangian.
I don't see a way to cancel out theta without cancelling out the ##\ddot{x}##, unless I am misunderstanding your suggestion.

TSny
Homework Helper
Gold Member
As you noted in post #3, ##\theta ## is proportional to ##x##. So, ##\ddot \theta## can be expressed in terms of ##\ddot x##.

Abourque72
Ah, yes. I get
$$\ddot{x} = \frac{-2kx}{m+ \frac{2kr}{g}}$$ which is the (hopefully) correct SHO which I can get the period from. Interesting to see that "correcting" mass term of ##\frac{2kr}{g}##; I wonder what kind of physical interpretation it entails.

TSny
Homework Helper
Gold Member
Let ##T## be the period of the motion of the system for small angles.

Let ##T_{Pend}## be the period of the pendulum alone. That is, it's the period of the pendulum if the point of attachment ##p## in your diagram were held fixed but the pendulum is free to rotate about ##p##.

Let ##T_{Spring}## be the period if the pivot at ##p## were frozen so that the pendulum cannot swing but ##p## is free to slide horizontally. So, the system just oscillates horizontally due to the springs while ##\theta## remains constant.

It is interesting to express ##T## in terms of ##T_{Pend}## and ##T_{Spring}##.

Abourque72
It is interesting to express ##T## in terms of ##T_{Pend}## and ##T_{Spring}##.

Indeed: ##T^2 = {T_{Pend}}^2 + {T_{Spring}}^2.##

TSny