Simple harmonic motion

In a nutshell

Simple harmonic motion is when an object oscillates either side of an equilibrium point. The displacement is defined as the distance from the equilibrium point and the acceleration is always directly proportional to the displacement and towards the midpoint.  

Equations

Variable definitions

Simple harmonic motion

Simple harmonic motion is defined as an object oscillating either side of a central equilibrium point. The displacement of an object in simple harmonic motion is given as the distance from the position of the object to the equilibrium.

As the object moves away from the equilibrium point, there is a restoring force which is either pushing or pulling the object towards the equilibrium point. The size of the force depends on the displacement of the object. 

Due to Newton's second law of motion, the restoring force always makes the object accelerate towards the midpoint, or equilibrium.  

Conditions for simple harmonic motion

There are two points you need to remember:

• The acceleration of the object is directly proportional to its displacement from the midpoint.
  
• The acceleration is always directed towards the midpoint.

Frequency and time period

The time period of an oscillation is measured from a maximum positive displacement, to a maximum negative displacement and back again and is measured in seconds $$s$$.​

The frequency is the number of oscillations every second and is measured in hertz $$Hz$$.

In simple harmonic motion, the frequency and time period of an oscillation are independent of the amplitude of the oscillation. This means that no matter how big the initial displacement of the oscillation is, the frequency and time period is constant. 

Example

This is why a pendulum clock will keep ticking at regular intervals even though over time the swing amplitude decays.

Curiosity: Simple harmonic motion is known as isochronous, meaning same time.

Simple harmonic equations

For an object oscillating horizontally, the motion can be transposed onto a circle and a displacement time graph.

The object completes one complete revolution of the circle for one oscillation. This means that the object will have an angular velocity $$\omega$$. 

Note: Angular frequency and angular velocity are interchangeable and mean the same thing, exams could refer to either. Both are represented as $$\omega$$.

To find the displacement of the object at any given time $$t$$ we can use the circle. Consider the following diagram. 

The horizontal displacement of the object from the midpoint is given as $$x$$ and the distance from the midpoint to the object is given as the radius of the circle $$r$$. The angle made from the horizontal displacement to the object is given as $$\theta$$.

From circular motion, angular velocity is given as:

​$$\omega =\frac \theta t$$​​

Rearranging for $$\theta$$:

$$\theta=\omega t$$    (1)​

Using trigonometry (where $$r$$ is the hypotenuse and $$x$$​ is the adjacent to the angle) to find an expression for the horizontal displacement $$x$$:

$$\cos \theta =\frac x r \newline \\[0.1in]x= r \cos \theta \newline \\[0.1in]$$    (2)​

Substituting $$\theta$$ from equation (1) into equation (2):

​$$x=r \cos (\omega t)$$​​

Lastly, the radius of the circle is given as the maximum horizontal displacement, which is called the amplitude of the oscillation, $$A$$:

$$x=A \cos (\omega t)$$​​

Note: This equation is also given as $$x=A \sin (\omega t)$$, which one you use depends on when the timing started. If the timing starts when the oscillation is at a maximum displacement then the $$\cos$$ version is used. If the timing starts when the oscillation is at the midpoint and therefore zero displacement $$\sin$$ is used. 

The velocity of the object $$v$$ at any displacement $$x$$ is given as: 

​$$v=\pm \omega \sqrt {(A^2-x^2)}$$​​

The acceleration of the object is directly proportional to the displacement, with $$-\omega^2$$ as the constant of proportionality:

$$a=-\omega^2x$$​​

Note: The negative sign is because the acceleration always acts in the opposite direction of the displacement.

​​Example

A pendulum bob is raised to a displacement of $$5.8 \, cm$$ from the midpoint. The bob is released and is measured to have a time period of $$4 \, s$$. Calculate its displacement at a time of $$0.68 \,s$$.

Firstly, write down the known values: 

$$T= 4 \, s \newline \\[0.1in]A=5.8 \, cm = 0.058 \,m$$

​Next, write down the equations needed and rearrange if necessary:

$$x=A\cos(\omega t) \space \space (1) \newline \\[0.1in] \omega = \dfrac {2 \pi}{T} \space \space (2)$$

Substitute $$(2)$$ into $$(1)$$:

$$x=A \cos (\frac {2\pi}{T} t)$$​​

Then, substitute the values into the equation:

$$x=0.058 \cos (\frac {2\pi}{4} \times 0.68) \newline \\[0.1in]x=0.02794...$$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$displacement \ x=0.03\ m$$​​​​​

The displacement of the pendulum is $$\underline{0.03\, m}.$$​

​

Maximum values

The maximum possible value that acceleration can take is when the displacement is the largest. This will happen when the displacement $$x$$​ is equal to the amplitude $$A$$:

$$a_{max}=\omega^2 A$$​​

The maximum possible value that velocity can have is when passing through the midpoint, when the displacement $$x$$ is $$0$$:

$$v_{max}=\omega \sqrt {A^2-0^2} \newline \\[0.1in]v_{max}=\omega \sqrt {A^2}\newline \\[0.1in]v_{max}=\omega A$$​​

Note: Make sure your calculator is in radians for all things circular motion and oscillations!

​​Example

A child is being pushed on a swing. The maximum displacement of the swing is $$2.40 \, m$$. If the time period of the swing is $$2.96 \, s$$, calculate the maximum velocity of the child.

Firstly, write down the known values: 

$$A=2.40 \, m \newline \\[0.1in]T=2.96 \, s$$

​Next, write down the equations needed and rearrange if necessary:

$$v_{max}=A\omega \space \space (1) \newline \\[0.1in]\omega = \dfrac {2 \pi}{T} \space \space (2)$$

Substitute $$(2)$$ into $$(1)$$:

$$v_{max}=A(\frac {2\pi}{T})$$​​

Then, substitute the values into the equation:

$$v_{max}=2.40 \times (\frac {2\pi}{2.96}) \newline \\[0.1in]v_{max}=5.09447...$$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$maximum \ velocity, \ v_{max}=5.09\ ms^{-1}$$​​​​​

The maximum velocity of the child is $$\underline{5.09\ ms^{-1}}.$$​

​