Equation of motion of a simple pendulum

[JD_PM]

Homework Statement

None

Relevant Equations

$$\ddot \theta + \frac{g}{l} \theta = 0$$

The equation of motion of a simple pendulum is:

$$\ddot \theta + \frac{g}{l} \theta = 0$$

Our Physics professor told us: 'If you want to become a good Physicist you have to be able to analytically check your answers to see whether they make sense'.

In class he took the limits of constant quantities of the problem, like the mass, to see what would physically happen.

However, here we are dealing with SHM, which doesn't depend on the mass of the oscillating object.

Thus, I only see two quantities we can play with: the length $l$ and gravity.

Analytical checking of length $l$

$$lim_{l -> \infty} = \frac{g}{l} = 0$$Then:

$$\ddot \theta = 0$$OK, so if the length of the rope is infinitely large, we expect no SHM at all. But how to justify it?

What I think is that, as the rope approaches infinity (and the angle is very small), we can think of the rope as being in the vertical position. That's the equilibrium position,in which the restoring force (i.e. the $\frac{g}{l}$ term) is zero.Analytical checking of gravity

$$lim_{g -> \infty} = \frac{g}{l} = \infty$$

In this case we have an infinite restoring force, which means that in absence of any kind of friction the mass will swing infinitely many times.

Do these analytical checks make sense to you?

 

[PeroK]

That's perhaps not the most appropriate analysis in this case. Perhaps looking at how energy relates to the two quantities might be better.

 

[JD_PM]

That's perhaps not the most appropriate analysis in this case. Perhaps looking at how energy relates to the two quantities might be better.

Mmm so you suggest to focus on energy. Let's try it.

The energy is conserved, so:

$$\Delta E = \frac{1}{2} m(l \dot \theta)^2 + mgl \cos \theta = 0 \ \ \ \ (1)$$

Where I took the zero gravitational potential line to be located at the pivot.

Equation (1) leads to

$$\dot \theta^2 = \frac{-2g}{l} \cos \theta$$

But here I'd tend to approach it the same way (taking limits)...

Is that what you meant?

 

[PeroK]

Is that what you meant?

Not really. We start with the differential equation for small oscillations:

The equation of motion of a simple pendulum is:

$$\ddot \theta + \frac{g}{l} \theta = 0$$

That equation leads to:

$\theta = \theta_0 \cos(\sqrt{\frac{g}{l}}t)$

Hence the period of the pendulum is:

$T = 2\pi\sqrt{\frac{l}{g}}$

We draw three conclusions:

1) The period is independent of the amplitude (for small amplitudes);
2) The period is proportional to the square root of the length;
3) The period is inversely proportional to the square root of $g$.

Can we justify or sanity-check these conclusions?

If we take $l$ to be very large, then the period is very large. I don't see that tells us much. And, for small $l$ the period becomes very small. I don't see that tells us much either.

If we take gravity to be very large, then the period becomes very small. That makes sense, as there is a strong restoring force.

If we take gravity to be small, then the period is large and oscillations are slow. Now, to me, that does make a lot of sense.

I would say we still need to justify 1) and 2) and we need to justify the square root in 3).

That's where perhaps considerations of energy (potential and kinetic) come into play?

 

[JD_PM]

If we take $l$ to be very large, then the period is very large. I don't see that tells us much. And, for small $l$ the period becomes very small. I don't see that tells us much either.

I've been trying to picture this situation in my mind and what I think is that, when $l$ is very large and the angle very small, $l$ has a very large vertical component in comparison to its horizontal component. Thus, the pendulum can be thought to be really close to its equilibrium position and its frequency is thus very small.

As the period is the inverse of the frequency, we of course expect the later to be really large.

This makes sense to me. It'd be nice to read your thoughts on this.

I would say we still need to justify 1) and 2) and we need to justify the square root in 3).

That's where perhaps considerations of energy (potential and kinetic) come into play?

Let me tackle 2) first.

From conservation of energy one gets (let me this time take the zero gravitational potential line to be located at the lowest point):

$$\Delta E = \frac{1}{2} m(l \dot \theta)^2 - mgl \cos \theta = 0 \ \ \ \ (2)$$

Then:

$$l \dot \theta = \sqrt{2gl \cos \theta}$$

We know that:

$$v = l \dot \theta$$

So

$$v= \sqrt{2gl \cos \theta}$$

So what we get is that the linear velocity is proportional to $\sqrt{gl}$.

I do not see how to justify that the period is proportional to the square root of the length by only using the conservation of energy approach though.

 

[PeroK]

What I was thinking of is:

The energy of the system is given by $E = mgh$ where $h$ is the initial vertical displacement of the pendulum. In this case $h = l(1 - \cos \theta_0) \approx \frac l 2 \theta_0^2$. So we have:

$E = \frac{mgl}{2}\theta_0^2$

And then do a bit of kinematic analysis with that.

 

[JD_PM]

I think I've got the idea, thanks.