# Energy of a Simple Harmonic Oscillator

### Energy changes in according to displacement

The kinetic energy of a simple harmonic oscillator is

\begin{aligned} E_\text{kinetic} &= \frac{1}{2}m \omega^2(x_o^2 - x^2) \end{aligned}

In a SHM, the oscillator’s kinetic energy and potential energy always changes from maximum to zero throughout the oscillations. However, at all time, the total energy of the oscillator is constant. This value can be obtained by calculating the maximum kinetic energy of the system:

\begin{aligned} E_\text{total} &= \frac{1}{2}mv_o ^2\\&=\frac{1}{2}m \omega^2 x_o^2 \end{aligned}

To find out the potential energy of a simple harmonic oscillator,

\begin{aligned} E_\text{potential} &= E_\text{total} - E_\text{kinetic}\\&=\frac{1}{2}m \omega^2x_o^2 - \frac{1}{2}m \omega^2(x_o^2 - x^2)\\&=\frac{1}{2}m\omega^2 x^2\end{aligned}

The energy changes of an SHM oscillator changes in a sinusoidal pattern.

### Energy changes according to time

From the velocity equation, the kinetic energy is

\begin{aligned} E_\text{kinetic}&=\frac{1}{2}mv^2\\&=\frac{1}{2}mv_o ^2 \cos^2{\omega t}\end{aligned}

The potential energy is the difference between the total energy and the kinetic energy,

\begin{aligned} E_\text{potential}&=\frac{1}{2}mv_o ^2 - \frac{1}{2}mv_o ^2 \cos^2{\omega t}\\&=\frac{1}{2}mv_o^2(1-\cos^2{\omega t})\\&=\frac{1}{2}mv_o \sin^2{\omega t} \end{aligned}

# The Physics of Simple Harmonic Motion

An object moving in an SHM must obey the relation

\begin{aligned} a=-\omega ^2 x\end{aligned}

where $x$ is the displacement and $\omega$ is the angular frequency of the oscillator. $\omega$  is related to the frequency $f$ of the oscillator by the relation \begin{aligned} w=2\pi f \end{aligned} .

A body is moving in SHM if its acceleration is directly proportional to the displacement and in the opposite direction to the displacement.

In this equation, we observe the following:

• the acceleration of the oscillator is always opposite to the displacement,
• the magnitude of the acceleration is directly proportional to the displacement.

From these two points, we see that the acceleration is greatest when the oscillator is at the maximum displacement while it is zero when the oscillator is at equilibrium. Since force and acceleration is directly proportional $F=ma$ , we see that the restoring force of the oscillator is correspondingly largest at the maximum displacement and zero at equilibrium.

### SHM in a spring

When a object hung on a spring oscillates, it is moving in SHM. There is a restoring force from the spring causing the object to move up and down. There are two ways to start the motion:

1. displace the object to a certain distance and release, or
2. with the oscillator at equilibrium, give it a push such that it starts moving.

Which colour of the graphs would case 1 and 2 belong?

The red graph would have an equation of \begin{aligned} x=x_0 \sin\omega t \end{aligned} while the blue graph is \begin{aligned} x=x_0 \cos \omega t \end{aligned}.

For the rest of this post, I will use the first equation (red), \begin{aligned} x = x_0 \sin \omega t \end{aligned}.

Since velocity is the derivative of displacement with time,

\begin{aligned} v=\frac{dx}{dt}=x_0 \omega \cos \omega t \end{aligned}

Acceleration is the derivative of velocity with time, hence

\begin{aligned} a=\frac{dv}{dt}= x_0 \omega^2 \sin \omega t\end{aligned}

But from the first equation $x = x_0 \sin \omega t$ , we have

\begin{aligned} a = -x_0 \omega^2 \end{aligned}

### Equation relating velocity to displacement

In the above equations, we have relationships with respect to time. We can also determine the velocity of the object if the displacement is known.

From the original displacement-time equation,

\begin{aligned} x&=x_0 \sin{\omega t} \end{aligned}

\begin{aligned} v=x_0 \omega \cos{\omega t} \end{aligned}

Squaring both equations, we have

\begin{aligned} x^2&=x_0 ^2 \sin^2{\omega t}\\v^2&=x_0 ^2 \omega ^2\cos^2{\omega t}\\&=x_0 ^2 \omega^2 (1-\sin^2{\omega t})\\&=x_0 ^2 \omega^2(1-\frac{x^2}{x_0 ^2})\end{aligned}

Simplifying,

\begin{aligned} v^2 &= \omega^2(x_0 ^2 - x^2) \end{aligned}

Hence,

\begin{aligned} v = \pm\omega \sqrt(x_0 ^2 - x^2) \end{aligned}

### The Restoring Force

It is important to understand that the restoring force is the resultant of both the spring force and the weight of the oscillator. Since weight is constant, the restoring force would be proportional to the spring force, $F=-kx$. Hence, the restoring force is never constant.

\begin{aligned} F_R=F_S-w \end{aligned}

where $F_R$ is the restoring force and $F_S$ is the spring force.

From the equation, it can be seen that the maximum restoring force occurs when $F_S$ is maximum, which occurs when the displacement is at maximum. This is also the position where the acceleration is maximum(and zero velocity).

### Phases of an SHM

\begin{aligned} x &= x_0 \sin \omega t \\ v &= x_0 \omega \cos \omega t\\ a &= -x_0 \omega ^2 \sin \omega t\end{aligned}

As we can see,

• the velocity of SHM $v$ and displacement $x$ has a phase difference of \begin{aligned} \frac{\pi}{2} \end{aligned} .
• The acceleration $a$ and the displacement has a phase difference of $\pi$.
• The acceleration and velocity has a phase difference of $\frac{\pi}{2}$

### Review Question 1

Using the blue graph above, sketch the corresponding graphs of $v$ and $a$.

# Simple Harmonic Motion and Circular Motion

Simple harmonic motion is closely related to a uniform circular motion.

A real life simple harmonic motion.

### Review Question 1

Answer the following questions as you watch the first YouTube video(Simple harmonic motion and uniform circular motion).

1. How does the centre of the red cylinder and the green ball relate?
2. At which part of the SHM(left) is the red cylinder moving the fastest and slowest?
3. At which part of the circular motion(right) is the green ball moving vertically the fastest and slowest?

### Review Question 2

Navigate to this site. This animation shows how a 4-stroke engine internal combustion works.

Answer the following questions:

1. The piston is moving with SHM. At which point is the piston the fastest?
2. At which point is the piston moving the slowest?

Given that the maximum velocity of the piston is $v$,

1. write down the expression of the velocity of the piston at the position shown in the diagram.
2. write down the expression of the acceleration of the piston at the position shown in the diagram.
3. Sketch a velocity-time graph of the SHM demonstrated by the piston. Draw two complete cycles and label the parts A, B and C on your graph.