# Simple Harmonic Motion

## Simple Harmonic Motion Revision

**Simple Harmonic Motion**

In this section we begin looking at objects in **simple harmonic motion (SHM)**. Examples of SHM can be seen around us from **pendulums** in clocks to a swing moving backwards and forwards.

**SHM**

**An oscillator is considered to be in simple harmonic motion (SHM) if the acceleration is proportional and opposite in direction to the displacement of the oscillator.**

The **acceleration** of the oscillator always acts in the same direction as the **restoring force**. This is the force that brings the oscillator back towards the **equilibrium position**. The acceleration can be calculated using the equation:

a=-\omega^2 x

- a=
**acceleration**in metres per second squared \text{(ms}^{-2}\text{)} - \omega=
**angular frequency**in radians per second (\text{rads}^{-1}) - x= the
**displacement**in metres \text{(m)}

The equation shows that the **maximum acceleration** occurs when the **displacement** is maximum.

**Example: **A simple harmonic oscillator has a time period of 2 \text{ s} when its maximum displacement is 0.05 \text{ m}. Calculate the maximum acceleration.

**[3 marks]**

\bold{a= -\omega^2 x} and \bold{\omega = \dfrac{2\pi}{T}}. By substitution:

\begin{aligned} a &= -\dfrac{4\pi^2}{T^2}x \\ &= -\dfrac{4\pi^2}{\textcolor{10a6f3}{2}^2} \times \textcolor{00d865}{0.05} \\ &= \bold{-0.5} \textbf{ms}\bold{^{-2}} \end{aligned}

The **displacement** is directly proportional to the **negative acceleration** of the simple harmonic oscillator. Plotting a displacement-acceleration graph forms a straight line through the origin where the gradient is equal to \omega^2.

From the graph we can see the points of **maximum displacement** at either end. These are known as the **amplitude** of the simple harmonic oscillator (A). The **displacement** (x) at any given point can be found using the equation:

x=ACos(\omega t)

- x= the
**displacement**in metres \text{(m)} - A= the
**amplitude**in metres \text{(m)} - \omega=
**the angular frequency**in radians per second \text{(rads}^{-1}\text{)} - t= the
**time**in seconds \text{(s)}

We can also calculate the **speed** at any given point of the simple harmonic oscillator. The speed of the oscillator would be at a minimum at its positive and negative amplitudes and at a maximum as it passes the equilibrium where x=0.

v=\pm \sqrt{A^2-x^2}

- v= the
**velocity**in metres per second \text{(ms}^{-1}\text{)} - A= the
**amplitude**in metres \text{(m)} - x= the
**displacement**in metres \text{(m)}

**Example: **A simple pendulum oscillates with simple harmonic motion with an amplitude of 0.3 \text{ m}. The frequency of the oscillations is 5 \text{ Hz}. Calculate the speed of the pendulum at a position of 0.1 \text{ m} from the equilibrium position.

**[3 marks]**

\bold{v= \pm \omega \sqrt{A^2-x^2}} and \bold{\omega = 2 \pi f}. So by substitution:

\begin{aligned} v &= \pm 2\pi f \sqrt{A^2-x^2} \\ &= \pm 2\times \pi \times \textcolor{00d865}{5} \times \sqrt{\textcolor{ffad05}{0.3}^2-\textcolor{10a6f3}{0.1}^2} \\ &= \bold{\pm 8.9}\textbf{ ms}\bold{^{-1}} \end{aligned}

**SHM Graphs**

Provided a simple harmonic oscillator is **undamped**, we should expect to see graphs similar to the ones below for any object on simple harmonic motion.

Displacement-time graph:

The x-t graph above is a simple sinusoidal graph. The amplitude (maximum displacement) always stays the same as there is no energy lost or gained during oscillations. Calculating the gradient at any point of the displacement-time graph gives the velocity.

Velocity-time graph:

The v-t graph above is a simple cosine graph. When displacement is zero from the previous graph, the velocity is maximum. When the oscillator is at its maximum displacement, the velocity is zero. Again, as no energy is gained or lost, the maximum velocity with each oscillation remains the same. Calculating the gradient at any point on the velocity-time graph gives acceleration.

Acceleration-time graph:

Calculating the **maximum velocity** of a simple harmonic oscillator can be done using a simpler equation than that learnt previously. As maximum velocity occurs when **displacement** (x) is zero, the equation can be simplified:

\begin{aligned} v &= \pm \omega \sqrt{A^2-x^2} \\ &= \pm \omega \sqrt{A^2+0^2} \\ &= \pm \omega \sqrt{A^2} \end{aligned}

v_{\text{max}}=\omega A

Calculating maximum acceleration can also be calculated, using the displacement at its maximum. Maximum displacement is known as the **amplitude**. Therefore the equation changes to:

a_{\text{max}}=-\omega A

**Mass-spring Systems**

A mass and a spring can form a system which moves in simple harmonic motion (SHM). If a mass is pulled to **maximum displacement **on a spring, a **restoring force** will return the mass to the **equilibrium position**.

The **time period **of the **mass-spring system **can be calculated using the equation:

T=2\pi \sqrt{\dfrac{m}{k}}

- T= the
**time period**in seconds \text{(s)} - m= the
**mass**in kilograms \text{(kg)} - k= the
**spring constant**in newtons per metre \text{Nm}^{-1}

The equation shows that the **time period** is proportional to **mass** and therefore, the greater the mass the greater the time period. It also shows an inversely proportional relationship between time period and **spring constant**.

**Example:** A spring with a spring constant of 2.2 \text{ Nm}^{-1} is extended by a mass of 5 \text{ kg}. The spring is stretched until it moves into simple harmonic motion. Calculate the time period of the oscillation.

**[2 marks]**

\begin{aligned} \bold{T} &= \bold{2\pi \sqrt{\dfrac{m}{k}}} \\ &= 2\pi \times \sqrt{\dfrac{\textcolor{f95d27}{5}}{\textcolor{aa57ff}{2.2}}} \\ &= \bold{9.5} \textbf{ s} \end{aligned}

**Simple Pendulums**

A mass and a string can form a **pendulum system **which moves in simple harmonic motion (SHM). If a mass is pulled to **maximum displacement **on a string, a **restoring force **will return the mass to the equilibrium position. The restoring force is the force responsible for bringing the oscillating object back to the equilibrium position.

The **time period **of the pendulum can be calculated using the equation:

T=2\pi\sqrt{\dfrac{L}{g}}

- T= the
**time period**in seconds \text{(s)} - L= the
**length of the string**in metres \text{(m)} - g= the
**gravitational field strength**in newtons per kilogram \text{(Nkg}^{-1}\text{)}

g is used in the equation above as this represents the** restoring force**. The equation shows that the time period is proportional to length and therefore, the longer the string the greater the time period.

**Example: **Calculate the time period of a pendulum of length 0.5 \text{ m} on Earth.

**[2 marks]**

\begin{aligned} \bold{T} &= \bold{2\pi \sqrt{\dfrac{L}{g}}} \\ &= 2\pi \times \sqrt{\dfrac{\textcolor{00d865}{0.5}}{9.81}} \\ &= \bold{1.42} \textbf{ s} \end{aligned}

**Required Practical 7**

**Calculating the spring constant of a mass-spring system.**

**Doing the experiment:**

- Set up the equipment as shown in the diagram.
- Begin with
**zero mass**on the mass hanger. This will give us a reading with the mass set to the mass of the mass hanger. - Pull the mass hanger down 5 \text{ cm} ensuring it passes the marker.
- Release the mass hanger and start the
**stopwatch**when it passes the marker the first time. The mass will**oscillate vertically**. - Stop the stopwatch after it passes the marker 10 times. Divide this time by 10 to get the time period (T) of the oscillations.
- Add 50 \text{ g} to the mass hanger and repeat the above steps.
- Continue this until you have 10 sets of results ranging in mass.

**Analysing the Results:**

Plot a graph of T^2 against m. You should get a graph that looks similar to the graph on the right hand side (a **direct proportionality**).

As the equation T = 2 \pi \sqrt{\dfrac{m}{k}} can be rearranged to give T^2 = 4 \pi ^2 \dfrac{m}{k}, the gradient of the graph represents \dfrac{4 \pi ^2}{k}.

Therefore k = \dfrac{4 \pi ^2}{\text{gradient}} where k is the **spring constant** of the spring.

## Simple Harmonic Motion Example Questions

**Question 1:** Describe how you would calculate the velocity of a simple harmonic oscillator from a displacement-time graph when the graph forms a curve.

**[2 marks]**

The **gradient of a displacement-time graph gives us the velocity**. As the graph is a curved line, you would need to use a **tangent to the curve** then find the gradient of the tangent.

**Question 2:** A spring with a spring constant of 5.1 \text{ Nm} ^{-1} is extended by a mass of 4 \text{ kg}. The spring is stretched until it moves into simple harmonic motion. Calculate the time period of the oscillation.

**[2 marks]**

**Question 3: **Calculate the frequency of a pendulum of length 1.2 \text{ m} on a planet with gravitational field strength of \dfrac{1}{5} of Earth.

**[3 marks]**

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