C1-1. Simple Harmonic Motion

1. Describing Oscillations & Defining SHM

An oscillation is the repetitive variation with time ($t$) of the displacement ($x$) of an object about its fixed equilibrium position ($x = 0$).

Key Terminology:

  • Amplitude ($x_0$): The maximum absolute displacement from the equilibrium position (m).
  • Period ($T$): The time taken for one complete oscillation cycle (s).
  • Frequency ($f$): The number of complete oscillations per second (Hz).
  • Angular Frequency ($\omega$): The rate of change of angular displacement in radians per second (rad s⁻¹).
$$T = \dfrac{2\pi}{\omega} = \dfrac{1}{f} \quad \Rightarrow \quad \omega = \dfrac{2\pi}{T} = 2\pi f$$

The Defining Condition of SHM:

Simple Harmonic Motion (SHM) occurs if and only if the acceleration ($a$) of the object is directly proportional to its displacement ($x$) and is directed entirely towards the equilibrium position (acting as a restoring force).

$$a = -\omega^2 x$$

Example 1

Problem: The period of a particle undergoing simple harmonic motion is $T$. What is the ratio of the acceleration of the particle to its displacement from the equilibrium position proportional to?

Solution:

  • Using the defining equation of SHM: $a = -\omega^2 x$.
  • Substitute $\omega = \dfrac{2\pi}{T}$: $$a = -\left(\dfrac{2\pi}{T}\right)^2 x = -\dfrac{4\pi^2}{T^2} x$$
  • Rearranging for the ratio: $$\dfrac{a}{x} = -\dfrac{4\pi^2}{T^2}$$
  • Conclusion: The ratio $\dfrac{a}{x}$ is universally proportional to $\mathbf{T^{-2}}$. This fundamentally proves that the period of SHM is entirely independent of the amplitude.

Example 2

Problem: A body undergoes one oscillation of simple harmonic motion. What is correct for the direction of the acceleration of the body and the direction of its velocity?

Solution:

  • When moving away from equilibrium, velocity points outward, but restoring acceleration points inward. They are in opposite directions (the object slows down). This happens for two separate quarter-periods ($\dfrac{1}{4}T \times 2$).
  • When moving towards equilibrium, velocity and acceleration both point inward. They are in the same direction (the object speeds up).
  • Conclusion: Acceleration and velocity are in opposite directions for exactly half a period.

Example 3

Problem: The graph shows the variation with time $t$ of the velocity $v$ of an object undergoing simple harmonic motion. At which velocity does the displacement from the mean position take a maximum positive value?

t v 0 A = π/2 B = π C = 3π/2 D = 2π

Solution:

Displacement reaches its absolute maximum positive value exactly when the velocity crosses zero after a positive half-cycle, so the distance takes a maximum positive value at D.

2. Kinematics of Simple Harmonic MotionHL Only

Because acceleration varies continuously with displacement, standard kinematic (SUVAT) equations completely fail in SHM. We model the motion using trigonometric functions.

Time-Dependent Kinematic Equations:

  • Starting at equilibrium ($x = 0$):
    $$ \begin{aligned} x &= x_0 \sin(\omega t) \\[0.1cm] v &= \omega x_0 \cos(\omega t) \end{aligned} $$
  • Starting at maximum displacement ($x = x_0$):
    $$ \begin{aligned} x &= x_0 \cos(\omega t) \\[0.1cm] v &= -\omega x_0 \sin(\omega t) \end{aligned} $$

Velocity-Displacement Equation:

To find the speed at a specific location without needing time $t$, we combine the trigonometric identities to yield:

$$v = \pm \omega \sqrt{x_0^2 - x^2}$$

Example 4: Equation of Motion and Phase Angle

Problem: A mass is released from a vertical height of $h_{\text{max}}$ at time $t = 0$. It oscillates with SHM of period $T$. (a) State the equation of motion. (b) A second mass-spring system oscillates with the same frequency but with a phase angle of $\phi = -\dfrac{\pi}{4}$. Describe the graph of the second system.

Solution:

  • Part A: Since it is released from maximum displacement, it follows a cosine curve. Substituting $\omega = \dfrac{2\pi}{T}$, the equation is: $$h = h_{\text{max}} \cos\left(\dfrac{2\pi}{T} t\right)$$
  • Part B: A phase angle of $-\dfrac{\pi}{4}$ indicates the wave is lagging. Geometrically, the entire graph is shifted to the right by $\dfrac{1}{8}$ of a full oscillation period (since $\dfrac{\pi}{4}$ is one-eighth of $2\pi$).