9.1.4 Examples of SHM

• Two examples of simple harmonic oscillators are:
  • A simple pendulum
  • A mass-spring system
• Considering equations related to the restoring force leads to expressions for the time periods of these scenarios
  • These relationships are useful for investigating simple harmonic motion experimentally

For a pendulum, the restoring force is provided by the component of the bob's weight that is perpendicular to the tension in the pendulum's string. For a mass-spring system, the restoring force is provided by the force of the spring

Time Period of a Simple Pendulum

• A simple pendulum is a type of simple harmonic oscillator
  • The pendulum consists of a string and a bob (a weight, generally spherical) at the end
• An oscillating pendulum can be modelled as simple harmonic motion when the angle of oscillation is small

Forces on a pendulum when it is displaced. Assuming θ < 10°, the small angle approximation can be used to describe the time period of a simple pendulum such as this.

• The restoring force, F, which returns an oscillating pendulum bob to the equilibrium position is:

F = −mg sin θ

• Where:
  • m = the mass of the pendulum bob (kg)
  • g = acceleration due to gravity (m s⁻²)
  • θ = angle between the bob and the vertical (°)

• Using Newton’s Second Law:

F = ma = −mg sin θ

• Both sides can be divided by m to give an expression for the acceleration, a:

a = −g sin θ

• In this case, the small-angle approximation can be used, this is where sin θ ≅ θ 
  • This is assuming the angle the pendulum makes with the vertical is less than 10°
• The expression for acceleration then becomes:

a = −gθ

• The displacement, x, is equal to the length of the arc made by the bob, x = Lθ
  • Where L = length of the pendulum (m)
• Rearranging this for θ and substituting it into the acceleration equation gives:

• This equation shows:
  • For small values of x, the condition for SHM is satisfied as restoring force, F is proportional to −x 
  • For large values of x, the acceleration of a simple pendulum is not proportional to the displacement

 

• Using the defining equation of SHM:

• Where ω = angular frequency of the oscillation (rad s⁻¹)
• Equating with the previously derived expression for acceleration leads to:

• The equation relating angular frequency, ω, and time period, T, is:

• Finally, combining these expressions leads to an equation for the time period of a simple pendulum:

Time Period of a Mass–Spring System

• A mass-spring system is another type of simple harmonic oscillator
• The restoring force, F, which returns a mass-spring system to its equilibrium position is given by:

F = −kx

• Where:
  • x = extension of the spring (m)
  • k = spring constant (N m⁻¹)

• Using Newton’s Second Law:

F = ma = −kx

• Rearranging for the acceleration, a:

• The defining equation of SHM is given by a = −ω²x

• Where:
  • m = mass (kg)
  • ω = angular frequency (rad s⁻¹)

• Combining these expressions for acceleration gives:

• This can be simplified to give an expression for ω:

• The equation relating angular frequency, ω, and time period, T, is:

• Finally, combining these expressions leads to an equation for the time period of a mass-spring system:

• This equation applies for both a horizontal or vertical mass-spring system

A mass-spring system can be either vertical or horizontal. The time period equation applies to both.

• The equation shows that:
  • The higher the spring constant k, the stiffer the spring and hence, the shorter the time period of oscillation
  • The time period (and hence, frequency) of a mass-spring system is independent of the force of gravity

• A consequence of this is that oscillations would have the same time period on Earth and the Moon

Worked Example

Step 1: List the known quantities

• • Length of the pendulum, L = 80 cm = 0.8 m
  • Acceleration due to gravity, g = 9.81 m s⁻²

Step 2: Write down the relationship between angular frequency, ω, and period, T

$\omega =\frac{2\pi }{T}$

$T=\frac{2\pi }{\omega }$

Step 3: Write down the equation for the time period of a simple pendulum

$T=2\pi \sqrt{\frac{L}{g}}$

• • This equation is valid for this scenario since the maximum angle of displacement is less than 10°

Step 4: Equate the two equations and rearrange for ω

$\frac{2\mathrm{\pi }}{\omega }=2\pi \sqrt{\frac{L}{g}}$

$\omega =\sqrt{\frac{g}{L}}$

Step 5: Substitute the values to calculate ω

$\omega =\sqrt{\frac{9.81}{0.8}}$= 3.50 rad s⁻¹

Step 6: State the final answer to the correct number of significant figures

ω = 3.5 rad s⁻¹

Worked Example

Step 1: Write down the known quantities

• • Mass, m = 2.0 kg
  • Spring constant, k = 0.9 N m⁻¹

Step 2: Write down the equation for the time period of a mass-spring system

Step 3: Write down the equation relating time period, T, and frequency, f

Step 4: Combine the equations and rearrange for frequency, f

Step 5: Substitute in the values to calculate frequency, f

= 0.1068 Hz

Step 6: State the final answer to the correct number of significant figures

f = 0.11 Hz

Exam Tip