Simple harmonic motion - Further Mathematics (9665) - Oxford AQA A-Level

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Simple Harmonic Motion (SHM): Comprehensive Study Notes (FM2: Mechanics)

Hello! Welcome to the exciting world of Simple Harmonic Motion. This chapter is fundamental to understanding how things oscillate—from the tiny vibrations in atoms to the massive swing of a wrecking ball. While the maths might look intimidating (we use second-order differential equations!), don't worry. If you master the definitions and the key formulae, the application becomes straightforward. Let's get moving!

1. Defining Simple Harmonic Motion (SHM)

The Core Condition

SHM describes the motion of an object where the acceleration is always directed towards a fixed point (the equilibrium position) and is directly proportional to the object's displacement from that point.

In simpler terms:

- The further you pull it away, the harder it pulls back.
- It always wants to return to the centre.

The Defining Differential Equation

The mathematical definition of SHM is given by a second-order differential equation relating displacement \(x\) and acceleration \(\frac{d^2x}{dt^2}\):

\[ \frac{d^2x}{dt^2} = -\omega^2 x \]

Key terms in this equation:

\(x\): The displacement from the equilibrium position (m).

\(\frac{d^2x}{dt^2}\): The acceleration of the particle (m s\(^{-2}\)).

\(-\) sign: This is crucial! It shows that the acceleration is always in the opposite direction to the displacement, meaning it points back towards \(x=0\).

\(\omega\): The angular frequency (rad s\(^{-1}\)). This is a positive constant that determines the speed of oscillation. Since \(\omega^2\) must be positive, any motion described by this equation must be SHM.

Analogy: Imagine a boat tied by a very stretchy rope to a fixed point on the shore. If the boat moves further out (\(x\) increases), the tension in the rope (force, and thus acceleration) increases, pulling it back towards the shore.

Quick Takeaway

To prove that a motion is SHM, you must show that the equation of motion derived from Newton's Second Law (\(F=ma\)) can be written in the form \(\frac{d^2x}{dt^2} = -(\text{positive constant}) \times x\). That positive constant is always \(\omega^2\).

2. Key Characteristics of SHM: Period, Frequency, and Amplitude

Once you find \(\omega\), you can determine all the key features of the motion.

a) Amplitude (\(A\))

The amplitude (\(A\)) is the maximum magnitude of displacement from the equilibrium position. It is the furthest distance the particle travels from the centre point.

- \(x\) always satisfies \(-A \leq x \leq A\).

b) Angular Frequency (\(\omega\))

We already met \(\omega\) in the defining equation. It links acceleration and displacement. Its value depends solely on the physical properties of the system (like mass and stiffness).

c) Period (\(T\)) and Frequency (\(f\))

The motion repeats itself over time. These properties describe how quickly that happens:

Period (\(T\)): The time taken for one complete oscillation or cycle (seconds).

\[ T = \frac{2\pi}{\omega} \]

Frequency (\(f\)): The number of oscillations completed per unit time (Hz or s\(^{-1}\)).

\[ f = \frac{1}{T} = \frac{\omega}{2\pi} \]

Did you know? High frequency means short periods (rapid shaking), while low frequency means long periods (slow swinging).

Quick Review Box

Relationships to Memorise:
- Defining Eq: \(\frac{d^2x}{dt^2} = -\omega^2 x\)
- Period: \(T = \frac{2\pi}{\omega}\)
- Frequency: \(f = \frac{\omega}{2\pi}\)

3. Velocity in SHM

For any given displacement \(x\), we can find the velocity \(v\) of the oscillating particle.

The Velocity Formula

The speed \(v\) at displacement \(x\) is given by the formula:

\[ v^2 = \omega^2 (A^2 - x^2) \]

Note: The syllabus uses 'a' for amplitude in the formula booklet, so you may see it written as \(v^2 = \omega^2 (a^2 - x^2)\). Be comfortable with both 'A' and 'a' representing amplitude.

Maximum and Minimum Speed

This formula immediately tells us where the particle is moving fastest and slowest.

Maximum Speed (\(V_{max}\)):

- Occurs when \(x = 0\) (at the equilibrium position).
- At \(x=0\), \(v_{max}^2 = \omega^2 A^2\), so \(V_{max} = A\omega\).

Minimum Speed (Zero Speed):

- Occurs when \(x = \pm A\) (at the endpoints of the motion).
- This makes sense: the particle must momentarily stop before reversing direction.

Common Mistake to Avoid: When using the velocity formula, remember that \(v\) represents speed if you take the positive root. If the question asks for velocity, you must include \(\pm\) and state the direction relative to \(x\). For example, at \(x > 0\), the particle could be moving towards or away from the equilibrium.

4. The Solution to the Differential Equation (Position in Time)

Solving the equation \(\frac{d^2x}{dt^2} = -\omega^2 x\) gives us an equation that tells us exactly where the particle is at any time \(t\).

General Solution Forms (FM2.6)

There are two primary forms for the solution you must be able to recognise and use:

Form 1: Using Phase Angle (\(\alpha\))

\[ x = A \cos(\omega t + \alpha) \]

Here, \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\alpha\) (alpha) is the phase angle or epoch. This angle relates to the starting position and velocity at \(t=0\).

Form 2: Using Two Constants (\(A\) and \(B\))

\[ x = A \cos(\omega t) + B \sin(\omega t) \]

Note: The constants \(A\) and \(B\) in this form are different from the amplitude \(A\) in Form 1. You determine these constants using the initial conditions (\(x\) and \(v\) at \(t=0\)).

Encouragement: Don't worry if this seems tricky at first. In practice, you often use one form or the other. Form 2 is sometimes easier if you need to differentiate to find velocity (\(v = \frac{dx}{dt}\)) as it avoids complex chain rule in the initial stages.

Connecting the Forms

If you have found the solution in Form 2, you can find the amplitude of the resulting motion (which is the \(A\) from Form 1) using basic trigonometry:

If \(x = A_1 \cos(\omega t) + B_1 \sin(\omega t)\), then the amplitude of the oscillation is \(R = \sqrt{A_1^2 + B_1^2}\).

The constants in the syllabus solution Form 2 are simply the coefficients of the cosine and sine terms that result from applying initial conditions.

Step-by-Step: Finding the Particular Solution

1. Find \(\omega\): Derive the equation of motion and write it as \(\frac{d^2x}{dt^2} = -\omega^2 x\).

2. Choose a Solution Form: Usually, \(x = A \cos(\omega t) + B \sin(\omega t)\) is convenient.

3. Find Velocity \(v\): Differentiate the chosen solution: \(v = \frac{dx}{dt}\).

4. Apply Initial Conditions (Boundary Values): Use the displacement and velocity at \(t=0\) (or any other known time) to find the values of \(A\) and \(B\).

5. Find Amplitude: If required, calculate the amplitude of the resulting motion using \(R = \sqrt{A^2 + B^2}\) (if you used Form 2).

5. Applications of SHM

SHM is used to model many systems in mechanics, primarily those involving springs and simple pendulums.

A) Mass on a Spring or Elastic String (FM2.6, referencing FM2.3)

For a particle of mass \(m\) oscillating under Hooke's Law, the force is \(F = -ke\), where \(e\) is the extension/compression and \(k\) is the stiffness.

1. Establish Equilibrium: When working with vertical springs, first find the natural extension (\(e_0\)) where the gravity force (\(mg\)) is balanced by the spring tension (\(ke_0\)).

2. Consider Displacement (\(x\)): Measure displacement \(x\) from this equilibrium point.

3. Apply Newton's Second Law (\(F=ma\)): The net restoring force is proportional to the displacement \(x\) from equilibrium. The resulting equation is:

\[ m\frac{d^2x}{dt^2} = -kx \]

4. Identify \(\omega\): Rearranging this gives \(\frac{d^2x}{dt^2} = -\frac{k}{m} x\). Therefore:

\[ \omega^2 = \frac{k}{m} \]

Where \(k = \frac{\lambda}{l}\) (Modulus of Elasticity \(\lambda\), original length \(l\)).

Key Takeaway for Springs: The angular frequency is determined by the ratio of stiffness to mass: \(\omega = \sqrt{\frac{k}{m}}\).

B) The Simple Pendulum

A simple pendulum consists of a point mass \(m\) suspended by a light inextensible string of length \(l\).

1. Forces: The restoring force that brings the mass back to the centre is the component of gravity acting perpendicular to the string, which is \(F = -mg \sin \theta\), where \(\theta\) is the angle of deflection.

2. Equation of Motion: Using Newton's Second Law (or rotational equivalent) for acceleration along the arc (\(a = l \frac{d^2\theta}{dt^2}\)):

\[ ml \frac{d^2\theta}{dt^2} = -mg \sin \theta \]

3. The Small Angle Approximation (Crucial for SHM):

True SHM only occurs if the angle \(\theta\) is small (typically less than 10 degrees). For small \(\theta\), we can use the approximation:

\[ \sin \theta \approx \theta \quad \text{ (where } \theta \text{ is in radians)} \]

4. The SHM Differential Equation: Applying the approximation and simplifying gives:

\[ \frac{d^2\theta}{dt^2} = -\frac{g}{l} \theta \]

5. Identify \(\omega\) and Period: This confirms the motion is SHM, with:

\[ \omega^2 = \frac{g}{l} \]

The Period \(T\) of a simple pendulum is:

\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{l}{g}} \]

Observation: Notice that the period of a simple pendulum does not depend on the mass \(m\) or the amplitude (as long as the angle is small). It only depends on the length \(l\) and gravity \(g\).

6. Summary and Final Tips

Essential Formulae Recap

If you know these four equations, you can solve most SHM problems:

1. Definition/Acceleration: \(\frac{d^2x}{dt^2} = -\omega^2 x\)
2. Velocity: \(v^2 = \omega^2 (A^2 - x^2)\)
3. Position (Form 2): \(x = A \cos(\omega t) + B \sin(\omega t)\)
4. Period: \(T = \frac{2\pi}{\omega}\)

Accessibility Tip: The Power of \(\omega\)

In every single problem, your first and most important goal is to find \(\omega\). Once you have \(\omega\), you can find everything else (Period, Frequency, \(v_{max}\)).

How to find \(\omega\)?

- For a spring: \(\omega = \sqrt{\frac{k}{m}}\) (or \(\sqrt{\frac{\lambda}{ml}}\)).
- For a pendulum: \(\omega = \sqrt{\frac{g}{l}}\).
- If the DE is given: \(\omega\) is the square root of the positive constant multiplying \(x\).

Final Encouragement: SHM beautifully connects mechanics and differential equations. Practice setting up the initial equation of motion correctly—this is where the marks are earned! You've got this!