Simple harmonic oscillations

1. Overview

Simple Harmonic Motion (SHM) is a specific, idealized form of periodic motion that serves as the foundation for understanding oscillations in physics. It occurs when a system is disturbed from a position of stable equilibrium and experiences a restoring force that attempts to return it to that equilibrium.

The fundamental physical principle governing SHM is that the acceleration of the oscillating object is directly proportional to its displacement from a fixed equilibrium position and is always directed towards that fixed position. This means the further the object is moved from the center, the harder it is "pulled" back. This relationship results in a characteristic sinusoidal motion where energy is continuously exchanged between kinetic and potential forms.

2. Key Definitions

To master SHM, you must use precise terminology. In the Cambridge 9702 exam, missing a single keyword (like "equilibrium" or "fixed point") can result in the loss of marks.

• Displacement ($x$): The distance of an object from its equilibrium position in a specified direction. It is a vector quantity. (Unit: $\text{m}$)
• Amplitude ($x_0$): The maximum displacement of an oscillating object from its equilibrium position. (Unit: $\text{m}$)
• Period ($T$): The time taken for an object to complete one full oscillation. (Unit: $\text{s}$)
• Frequency ($f$): The number of oscillations per unit time. It is the reciprocal of the period ($f = 1/T$). (Unit: $\text{Hz}$ or $\text{s}^{-1}$)
• Angular Frequency ($\omega$): A measure of the rate of rotation or oscillation expressed in radians per second. It relates the linear frequency to the phase of the motion. (Unit: $\text{rad s}^{-1}$)
• Phase Difference ($\phi$): The difference in the fraction of a cycle between two oscillating systems, or between two points in the same cycle, often measured in radians.
• Simple Harmonic Motion (SHM): The motion of an oscillator such that its acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.

3. Content

3.1 The Defining Equation of SHM

The definition of SHM can be summarized by a single mathematical condition. For any system to be considered a simple harmonic oscillator, it must obey: $$a \propto -x$$ When we introduce the constant of proportionality, we get the defining equation: $$\mathbf{a = -\omega^2 x}$$

Analysis of the Equation:

1. The Negative Sign: This is the most important conceptual part of the equation. It indicates that the acceleration $a$ and the displacement $x$ are always in opposite directions. If the object is displaced to the right (+), the acceleration acts to the left (-).
2. The $\omega^2$ Term: Since any real number squared is positive, $\omega^2$ ensures the proportionality constant is positive, maintaining the requirement that $a$ and $x$ have opposite signs.
3. The Fixed Point: The displacement $x$ is always measured from the equilibrium position (where $x=0$). At this point, the acceleration is also zero.

3.2 Relationships between $T, f,$ and $\omega$

Angular frequency $\omega$ links the timing of the oscillation to the geometry of a circle (as SHM can be viewed as a projection of uniform circular motion).

1. $f = \frac{1}{T}$
2. $\omega = 2\pi f$
3. $\omega = \frac{2\pi}{T}$

In calculations, if you are given the period $T$, your first step should almost always be to calculate $\omega$.

3.3 Kinematic Equations (Functions of Time)

The motion of an SHM oscillator is sinusoidal. The specific equation used depends on the starting position of the object at $t = 0$.

Displacement ($x$)

If the object starts at the equilibrium position ($x = 0$) at $t = 0$: $$\mathbf{x = x_0 \sin(\omega t)}$$ If the object starts at the maximum displacement ($x = x_0$) at $t = 0$: $$\mathbf{x = x_0 \cos(\omega t)}$$ (Note: The Cambridge syllabus primarily focuses on the sine version, but you should be comfortable with both.)

Velocity ($v$)

Velocity is the first derivative of displacement with respect to time ($v = \frac{dx}{dt}$). Using the sine displacement equation: $$v = \frac{d}{dt}(x_0 \sin(\omega t))$$ $$\mathbf{v = \omega x_0 \cos(\omega t)}$$ Since the maximum value of a cosine function is 1, the maximum velocity ($v_0$) occurs when the object passes through the equilibrium position: $$\mathbf{v_0 = \omega x_0}$$ This allows us to write: $v = v_0 \cos(\omega t)$.

Acceleration ($a$)

Acceleration is the derivative of velocity ($a = \frac{dv}{dt}$): $$a = \frac{d}{dt}(\omega x_0 \cos(\omega t))$$ $$\mathbf{a = -\omega^2 x_0 \sin(\omega t)}$$ Substituting $x = x_0 \sin(\omega t)$ back into this expression gives the defining equation: $a = -\omega^2 x$. The maximum acceleration ($a_0$) occurs at the amplitude (maximum displacement): $$\mathbf{a_0 = \omega^2 x_0}$$

3.4 Velocity in terms of Displacement

Often, you need to find the speed of an oscillator at a specific point in its path without knowing the time. We derive this using the identity $\sin^2\theta + \cos^2\theta = 1$.

1. Start with $x = x_0 \sin(\omega t) \implies \sin(\omega t) = \frac{x}{x_0}$
2. Start with $v = \omega x_0 \cos(\omega t) \implies \cos(\omega t) = \frac{v}{\omega x_0}$
3. Substitute into the identity: $\left(\frac{x}{x_0}\right)^2 + \left(\frac{v}{\omega x_0}\right)^2 = 1$
4. Rearrange for $v$: $$\frac{v^2}{\omega^2 x_0^2} = 1 - \frac{x^2}{x_0^2}$$ $$v^2 = \omega^2 x_0^2 \left(1 - \frac{x^2}{x_0^2}\right)$$ $$v^2 = \omega^2 (x_0^2 - x^2)$$
5. Final Equation: $$\mathbf{v = \pm \omega \sqrt{x_0^2 - x^2}}$$ The $\pm$ sign indicates that at any displacement $x$ (except at the amplitude), the object could be moving in either direction.

3.5 Graphical Representations

Variations against Time ($t$)

When plotting $x, v,$ and $a$ against time, notice the phase shifts:

• Displacement-Time ($x-t$): A sine wave.
• Velocity-Time ($v-t$): A cosine wave. It is $\pi/2$ rad ($90^\circ$) out of phase with displacement. Velocity is maximum when displacement is zero.
• Acceleration-Time ($a-t$): An inverted sine wave. It is $\pi$ rad ($180^\circ$) out of phase with displacement. Acceleration is maximum (in the negative direction) when displacement is at its positive maximum.

Variations against Displacement ($x$)

• Acceleration vs. Displacement ($a-x$): This is a straight line through the origin with a negative gradient.
  • The gradient of the $a-x$ graph is equal to $-\omega^2$.
• Velocity vs. Displacement ($v-x$): This forms an ellipse. Velocity is zero at $x = \pm x_0$ and maximum at $x = 0$.

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Worked Example 1 — Calculating Kinematic Variables

A horizontal platform oscillates with SHM in a vertical plane with a period of $0.80 \text{ s}$ and an amplitude of $5.0 \text{ cm}$. A small mass rests on the platform. Calculate the maximum acceleration of the mass and determine if it remains in contact with the platform.

Step 1: Convert units to SI $T = 0.80 \text{ s}$ $x_0 = 5.0 \text{ cm} = 0.050 \text{ m}$

Step 2: Calculate angular frequency ($\omega$) $$\omega = \frac{2\pi}{T}$$ $$\omega = \frac{2\pi}{0.80} = 7.854 \text{ rad s}^{-1}$$

Step 3: Calculate maximum acceleration ($a_0$) $$a_0 = \omega^2 x_0$$ $$a_0 = (7.854)^2 \times 0.050$$ $$a_0 = 61.685 \times 0.050 = 3.084 \text{ m s}^{-2}$$ Answer: $a_{max} = 3.1 \text{ m s}^{-2}$ (to 2 s.f.)

Step 4: Analyze contact For the mass to lose contact, the downward acceleration of the platform must exceed the acceleration due to gravity ($g = 9.81 \text{ m s}^{-2}$). Since $3.1 < 9.81$, the mass remains in contact throughout the oscillation.

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Worked Example 2 — Velocity at a Specific Displacement

An object undergoing SHM has a frequency of $5.0 \text{ Hz}$ and an amplitude of $12 \text{ mm}$. Calculate the speed of the object when it is $8.0 \text{ mm}$ from the equilibrium position.

Step 1: Identify variables and convert to SI $f = 5.0 \text{ Hz}$ $x_0 = 12 \text{ mm} = 0.012 \text{ m}$ $x = 8.0 \text{ mm} = 0.008 \text{ m}$

Step 2: Calculate $\omega$ $$\omega = 2\pi f = 2 \times \pi \times 5.0 = 10\pi \approx 31.42 \text{ rad s}^{-1}$$

Step 3: Use the velocity-displacement equation $$v = \pm \omega \sqrt{x_0^2 - x^2}$$ $$v = 31.42 \times \sqrt{0.012^2 - 0.008^2}$$ $$v = 31.42 \times \sqrt{0.000144 - 0.000064}$$ $$v = 31.42 \times \sqrt{0.00008}$$ $$v = 31.42 \times 0.008944$$ Answer: $v = 0.28 \text{ m s}^{-1}$ (to 2 s.f.)

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Worked Example 3 — Interpreting an $a-x$ Graph

A student plots a graph of acceleration $a$ against displacement $x$ for a pendulum. The graph is a straight line passing through $(0,0)$ and the point $(0.040 \text{ m}, -1.6 \text{ m s}^{-2})$. Calculate the period of the pendulum.

Step 1: Determine the gradient of the graph $$\text{Gradient} = \frac{\Delta a}{\Delta x} = \frac{-1.6 - 0}{0.040 - 0} = -40 \text{ s}^{-2}$$

Step 2: Relate gradient to $\omega$ In SHM, $a = -\omega^2 x$, so the gradient is $-\omega^2$. $$-\omega^2 = -40$$ $$\omega = \sqrt{40} = 6.325 \text{ rad s}^{-1}$$

Step 3: Calculate the period $T$ $$T = \frac{2\pi}{\omega}$$ $$T = \frac{2\pi}{6.325} = 0.993 \text{ s}$$ Answer: $T = 0.99 \text{ s}$ (to 2 s.f.)

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4. Key Equations

Equation	Description	Data Sheet?
$a = -\omega^2 x$	Defining equation of SHM	Yes
$\omega = 2\pi f = \frac{2\pi}{T}$	Angular frequency relations	Yes
$v = \pm \omega \sqrt{x_0^2 - x^2}$	Velocity as a function of displacement	Yes
$x = x_0 \sin(\omega t)$	Displacement as a function of time	No (Memorise)
$v = v_0 \cos(\omega t)$	Velocity as a function of time	No (Memorise)
$v_0 = \omega x_0$	Maximum velocity (at equilibrium)	No (Memorise)
$a_0 = \omega^2 x_0$	Maximum acceleration (at amplitude)	No (Memorise)

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5. Common Mistakes to Avoid

• ❌ Wrong Calculator Mode: Using Degrees instead of Radians when calculating $\sin(\omega t)$.
  • ✓ Right: SHM equations are derived from circular motion and calculus; $\omega t$ is an angle in radians. Always check for the "R" or "Rad" icon on your screen.
• ❌ Confusing $x$ and $x_0$: Using the amplitude ($x_0$) in the $a = -\omega^2 x$ equation when the question asks for acceleration at a specific point.
  • ✓ Right: $x$ is the instantaneous displacement; $x_0$ is the constant maximum displacement.
• ❌ Ignoring the Negative Sign: Treating $a = \omega^2 x$ as the definition.
  • ✓ Right: The negative sign is essential because it shows the restoring nature of the motion. Without it, the object would accelerate away from equilibrium forever.
• ❌ Unit Mismatch: Mixing $\text{cm}$ and $\text{mm}$ with $\text{m s}^{-1}$.
  • ✓ Right: Convert all distances to meters ($\text{m}$) at the start of the problem to ensure consistency with $g$ ($9.81 \text{ m s}^{-2}$) and other SI units.
• ❌ Phase Confusion: Thinking velocity is maximum at maximum displacement.
  • ✓ Right: At maximum displacement, the object momentarily stops ($v=0$). Velocity is maximum at the equilibrium position ($x=0$).

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6. Exam Tips

1. The "Two-Mark" Definition: If asked to define SHM, always write: "Acceleration is proportional to displacement AND directed towards a fixed point (or opposite to displacement)." You need both parts for full marks.
2. Graph Gradients:
   • The gradient of an $x-t$ graph is velocity.
   • The gradient of a $v-t$ graph is acceleration.
   • The gradient of an $a-x$ graph is $-\omega^2$.
3. Identifying SHM from a Graph: If you are shown a graph of $a$ vs $x$, it must be a straight line through the origin with a negative gradient to prove the motion is SHM.
4. Maximum Values: Questions often ask for "maximum speed." Remember this occurs at $x=0$. Use $v_0 = \omega x_0$. If they ask for "maximum acceleration," it occurs at $x=x_0$. Use $a_0 = \omega^2 x_0$.
5. Phase Difference Calculation: To find the phase difference $\phi$ between two waves on a time graph, use the formula: $$\phi = \frac{\Delta t}{T} \times 2\pi$$ where $\Delta t$ is the time shift between identical points on the two waves.