In harmonic motion there is always a *restorative force*, which acts in the opposite direction of the velocity. The restorative force changes during oscillation and depends on the position of the object. In a pendulum it is the component of gravity along the path of motion. The force on the oscillating object is directly opposite that of the direction of velocity. For pendulums, the period gets larger as the length of the pendulum increases.

\begin{align*} T_{\text{pendulum}} = 2\pi\sqrt{\frac{L}{g}}\end{align*} ; Period of a pendulum of length L

#### Example

You have a mass swinging on the end of 1 m pendulum. If the maximum linear velocity of the mass is 2 m/s, (a) calculate the period of the pendulum and (b) calculate the amplitude of the pendulum.

To calculate the period of the pendulum, we can just plug in the given length into the equation above.

\begin{align*} T&=2\pi\sqrt{\frac{l}{g}}\\ T&=2\pi\sqrt{\frac{1\:\text{m}}{9.8\:\text{m/s}^2}}\\ T&=2\:\text{s}\\ \end{align*}

To find the amplitude, we'll use the equation given in the Period and Frequency lesson that gives us the velocity as a function of time. Since the problem says that the given velocity is the maximum velocity, we know that the pendulum is at the bottom of it's arc and 1/4th (or 3/4th's) of it's way through one period. Based on this knowledge, we can plug in 1/4 of the period for the change in time. We also know the frequency because we just found the period, so all we have to do is solve for the amplitude.

\begin{align*} v(t)&=-2\pi fA\cos(2\pi f\Delta t) && \text{start with the equation for velocity}\\ v_{max}&=-2\pi fA\cos(2\pi f\frac{1}{4}T) && \text{put in the terms we know}\\ A&=-\frac{v_{max}}{2\pi f\cos(2\pi f\frac{1}{4}T)} && \text{solve for the amplitude}\\ A&=-\frac{2\:\text{m/s}}{2\pi \frac{1}{2}\:\text{Hz}*\cos(2\pi *\frac{1}{2}\:\text{Hz}*\frac{1}{4}*2\:\text{s})} && \text{plug in the known values}\\ A&=.63\:\text{m} \end{align*}

### Interactive Simulation

### Review

- Why doesn’t the period of a pendulum depend on the mass of the pendulum weight? Shouldn’t a heavier weight feel a stronger force of gravity?
- The pendulum of a small clock is \begin{align*}1.553\;\mathrm{cm}\end{align*} long. How many times does it go back and forth before the second hand goes forward one second?
- On the moon, how long must a pendulum be if the period of one cycle is one second? The acceleration of gravity on the moon is one sixth that of Earth.

### Review (Answers)

- A heavier weight feels a stronger force of gravity, but is also harder to accelerate.
- \begin{align*}4\end{align*} times
- \begin{align*}0.04 \;\mathrm{m}\end{align*}

### Explore More

We have explored two examples of simple harmonic motion: the pendulum and the mass-spring system in the previous lesson. The purpose of this investigation is to get you to notice the connections between the two systems. Your task: Match the period of the pendulum system with that of the spring system. You are only allowed to change the mass involved in the spring system. Consider the effective length of the pendulum to be fixed at 2m because that is the distance between the center of mass and the pivot. The spring constant(13.5N/m) is also fixed. You may use any relationships you have learned about to help you. You should view the charts to check whether you have succeeded.

Instructions: To alter the mass, simply click on the 'select' tool in the menu, and select the mass. Then use the tab at the bottom of your screen to change the density or dimensions of the block to get the mass that you want.

To view chart legend, click on Settings and you can plot velocity or position of the pendulum or mass on the spring.

The mass and the spring constant have now been changed. What is the new period of the mass-spring system? Can you change the length of the pendulum to match the periods now?