# Simple pendulum, changing amplitude vs changing the length of string

1. May 15, 2013

### Bengo

Hello,

I don't understand how changing the amplitude does not change the period but changing the length of the string does. Each changes the height and changes the arc length so wouldn't they both change the period?

Thank you for any help!

2. May 16, 2013

### Simon Bridge

Changing the length of the string only changes the arc-length if you keep the initial angle fixed.

Have you tred this out in an experiment? (Keeping the arc-length the same for different string-lengths?)
What you have noticed is that the period of the pendulum does not depend on the arc-length - therefore the mechanism by which changing the string-length changes the period does not involve the arc-length either.

For a bigger arc-length, the height dropped is bigger, so the speed at the bottom is faster, so the period remains unchanged. So you need to look more carefully at what is different for a longer pendulum.

3. May 16, 2013

### DEvens

A pendulum only works that way for small angles. The "buzz phrase" is simple harmonic motion.

Consider a system where the force trying to bring the object back to "zero" is proportional to the size of the displacement from zero. For example, a perfect spring with force constant K.

F = - K x

A pendulum will work that way for small angles, because the force towards zero angle will be proportional to the angle. That's some keen geometry homework. Show it's true because for small theta, sin(theta) is proportional to theta. And show what the effective spring constant K is, and so get an effective equation that looks like F = - K x.

So if F = - K x, then the second derivative w.r.t. time (the acceleration a) is proportional to x.

F = m a = - K x
so
a = (- K/m) x

And that can be solved exactly for x as a function of time. Suppose x(t=0) is D, and suppose speed at t=0 is 0. That is, we pull the thing back and let it go at t=0. Then you get x = D cos(w t), and w^2 = K/m. That's because the derivative of cos(w t) w.r.t. to time t is - w sin(w t). And the derive of that is -w^2 cos(w t).

But notice that w, the angular frequency, gives the period. The period is w/(2 pi). And note also that w only depends on K/m, not on D. That is, the amplitude does not affect the frequency.

So, for a pendulum, if the angle is small, then the frequency does not depend on the amplitude. And it's because the pendulum (approximately) satisfies an equation that looks like F = - K x.
Dan

4. May 16, 2013

### WannabeNewton

DEvens made the key point: what you are speaking of only works for very small amplitudes so if you wanted to keep the motion simple harmonic but wanted to change the amplitude you must change it by an infinitesimal amount, thus it wouldn't affect anything regardless. If we do not assume small oscillations, the equations of motion remain as $\ddot{\theta} + \sin\theta = 0$ (I have set $m = L = g = 1$ for simplicity). Hence the period of oscillation is given by $T = \frac{4}{\sqrt{2}}\int _{0}^{\theta_{\max}}\frac{d\theta}{\sqrt{\cos\theta - \cos\theta_{\max}}}$ which does depend on the amplitude $\theta_{max}$.