Is the Motion of a Pendulum Simple Harmonic?

[Calpalned]

Homework Statement

My textbook states that for oscillations of pendulums, the restoring force is $F = -mgsin(\theta)$. "Because F is proportional to the sine of $\theta$ and not $\theta$ itself, the motion is not SHM (simple harmonic motion)". I don't understand the last sentence.

Homework Equations

For small angles, $sin(\theta) ≈ \theta$.

The Attempt at a Solution

Why is it that if something is proportional to $\theta$ it is SHM, but $sin(\theta)$ is not SHM? What's the difference?

 

[SammyS]

Homework Statement

My textbook states that for oscillations of pendulums, the restoring force is $F = -mgsin(\theta)$. "Because F is proportional to the sine of $\theta$ and not $\theta$ itself, the motion is not SHM (simple harmonic motion)". I don't understand the last sentence.

Homework Equations

For small angles, $sin(\theta) ≈ \theta$.

The Attempt at a Solution

Why is it that if something is proportional to $\theta$ it is SHM, but $sin(\theta)$ is not SHM? What's the difference?

How simple, or how complex an answer do you want?

One feature of simple harmonic motion is that the motion is sinusoidal as a function of time. It's also true that the period id independent of the amplitude.

Neither of those is exactly true if $\ F = -mg\sin(\theta) \$ .

 

[andrevdh]

The derivation of the SHM theory usually starts out with something like

F = -kx ...

that is the restoring force is directly proportional to the displacement.
If the force is directly proportional to θ, and θ is small, the SHM theoretical equations
can again be derived, but not if F is proportional to the sine of theta. So in summary
the pendulum motion can not be descibed by the SHM equations for large amplitudes
and it is only an approximation for small amplitudes.