Chapter: Simple Harmonic Motion (SHM)
Welcome to one of the most elegant and practical topics in CCEA A2 Mechanics 1! Simple Harmonic Motion describes the rhythmic back-and-forth oscillations seen everywhere in the physical world — from the gentle swaying of a grandfather clock pendulum to the bouncing of a mass on a spring. While the mathematics involves differential equations and trigonometry, the underlying principles are logical, structured, and very manageable once broken down step by step.
Don't worry if this topic feels a bit abstract at first. By the end of this chapter, you will master the defining equations, learn how to calculate positions, speeds, and times, and apply these concepts to real mechanical systems like elastic strings and simple pendulums.
---1. The Core Definition and Governing Equation of SHM
What is Simple Harmonic Motion?
A particle is said to move with Simple Harmonic Motion (SHM) when it oscillates along a straight line about a fixed central equilibrium point \(O\), such that its acceleration is directly proportional to its displacement from \(O\) and is always directed towards \(O\).
Because the acceleration always points back towards the centre, it acts as a "restoring" effect. When the particle is to the right of the centre (\(x > 0\)), acceleration pulls it to the left (\(\ddot{x} < 0\)). When the particle is to the left (\(x < 0\)), acceleration pulls it to the right (\(\ddot{x} > 0\)).
The Differential Equation of SHM
Mathematically, we write this relationship as:
\(\ddot{x} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\omega^2 x\)
Where:
• \(x\) is the displacement from the fixed centre of oscillation \(O\) (measured in metres, \(\text{m}\)).
• \(\ddot{x}\) is the acceleration of the particle (measured in \(\text{m s}^{-2}\)).
• \(\omega\) is a positive constant known as the angular frequency (measured in radians per second, \(\text{rad s}^{-1}\)).
• The minus sign ensures that acceleration is always in the opposite direction to displacement (pointing back towards \(O\)).
Deriving the Velocity–Displacement Formula
In mechanics, acceleration can be written in two standard ways: \(\frac{\mathrm{d}v}{\mathrm{d}t}\) or \(v \frac{\mathrm{d}v}{\mathrm{d}x}\). Using the spatial derivative allows us to connect velocity \(v\) directly to position \(x\):
\(v \frac{\mathrm{d}v}{\mathrm{d}x} = -\omega^2 x\)
Separating variables and integrating both sides:
\(\int v \, \mathrm{d}v = -\omega^2 \int x \, \mathrm{d}x\)
\(\frac{1}{2}v^2 = -\frac{1}{2}\omega^2 x^2 + C\)
To find the constant of integration \(C\), we use the physical boundary condition: at the extreme edge of the motion (where displacement reaches its maximum amplitude \(x = a\)), the particle momentarily comes to rest (\(v = 0\)):
\(0 = -\frac{1}{2}\omega^2 a^2 + C \implies C = \frac{1}{2}\omega^2 a^2\)
Substituting \(C\) back into our equation gives the fundamental velocity-displacement relationship:
\(v^2 = \omega^2 (a^2 - x^2)\)
Key Takeaway: The defining condition of SHM is \(\ddot{x} = -\omega^2 x\). From this, the relationship connecting speed to position is \(v^2 = \omega^2(a^2 - x^2)\).
---2. Key Parameters and Maximum Values
Essential Quantities
• Amplitude (\(a\)): The maximum displacement of the particle from the centre \(O\). The motion takes place entirely in the interval \(-a \le x \le a\).
• Periodic Time (\(T\)): The time taken to complete one full oscillation (e.g., from \(O\) to \(+a\), back through \(O\) to \(-a\), and returning to \(O\)):
\(T = \frac{2\pi}{\omega}\)
• Frequency (\(f\)): The number of complete cycles per second (measured in Hertz, \(\text{Hz}\)):
\(f = \frac{1}{T} = \frac{\omega}{2\pi}\)
Maximum Speed and Acceleration
Understanding where speed and acceleration reach their peak values is crucial for exam problems:
• Maximum Speed (\(v_{\max}\)): Occurs when passing through the equilibrium centre (\(x = 0\)). Setting \(x = 0\) in the velocity formula gives:
\(v_{\max} = a\omega\)
• Maximum Acceleration (\(a_{\max}\)): Occurs at the extreme turnaround points (\(x = \pm a\)). Setting \(x = a\) in the magnitude of acceleration gives:
\(a_{\max} = a\omega^2\)
Analogy: Think of a playground swing. At the very bottom (centre of motion), you are moving at your fastest (\(v_{\max}\)) but feel no horizontal pulling force towards the centre. At the highest points of your swing (extremes), you stop completely for an instant (\(v = 0\)), but the restoring force pulling you back is at its maximum (\(a_{\max}\)).
Key Takeaway: Speed is greatest at the centre (\(v_{\max} = a\omega\)), while acceleration is greatest at the edges (\(a_{\max} = a\omega^2\)).
---3. Displacement–Time Solutions
Solving the differential equation \(\ddot{x} = -\omega^2 x\) with respect to time yields sinusoidal solutions. The choice of equation depends entirely on where the particle is located when timing starts (\(t = 0\)).
Case A: Timing Starts at the Centre (\(t = 0, x = 0\))
If the particle passes through the centre \(O\) moving in the positive direction at \(t = 0\):
\(x(t) = a \sin(\omega t)\)
Differentiating with respect to time gives the velocity:
\(v(t) = \dot{x} = a\omega \cos(\omega t)\)
Case B: Timing Starts at the Maximum Displacement (\(t = 0, x = a\))
If the particle is released from rest at the positive extreme position \(x = a\) at \(t = 0\):
\(x(t) = a \cos(\omega t)\)
Differentiating with respect to time gives the velocity:
\(v(t) = \dot{x} = -a\omega \sin(\omega t)\)
General Solution
For any arbitrary starting position, the displacement can be written as:
\(x(t) = A \cos(\omega t) + B \sin(\omega t) = a \sin(\omega t + \varepsilon) \quad \text{or} \quad a \cos(\omega t + \phi)\)
Memory Trick:
• Sin starts at the Spot in the middle (\(x = 0\)).
• Cos starts at the Crest / edge (\(x = a\)).
Worked Example 1: Finding Speeds and Times
Problem: A particle performs SHM about an origin \(O\) with an amplitude of \(4\text{ m}\) and a period of \(6\text{ s}\). The particle is released from rest at \(x = 4\text{ m}\) at time \(t = 0\).
Find:
(a) The value of \(\omega\).
(b) The speed of the particle when \(x = 2.5\text{ m}\).
(c) The time taken to travel directly from \(x = 4\text{ m}\) to \(x = 2.5\text{ m}\).
Solution:
(a) Using the period formula \(T = \frac{2\pi}{\omega}\):
\(6 = \frac{2\pi}{\omega} \implies \omega = \frac{2\pi}{6} = \frac{\pi}{3}\text{ rad s}^{-1}\)
(b) Using the formula \(v^2 = \omega^2(a^2 - x^2)\):
\(v^2 = \left(\frac{\pi}{3}\right)^2 (4^2 - 2.5^2) = \left(\frac{\pi^2}{9}\right)(16 - 6.25) = \left(\frac{\pi^2}{9}\right)(9.75)\)
\(v = \frac{\pi}{3}\sqrt{9.75} \approx 3.27\text{ m s}^{-1}\)
(c) Since timing starts at the extreme position \(x = 4\text{ m}\), use \(x = a \cos(\omega t)\):
\(2.5 = 4 \cos\left(\frac{\pi}{3} t\right)\)
\(\cos\left(\frac{\pi}{3} t\right) = \frac{2.5}{4} = 0.625\)
Ensure your calculator is in radian mode:
\(\frac{\pi}{3} t = \arccos(0.625) \approx 0.89566\text{ rad}\)
\(t = \frac{3 \times 0.89566}{\pi} \approx 0.855\text{ s}\)
Key Takeaway: Always select your displacement equation based on the boundary condition at \(t = 0\), and ensure your calculator is set to radians when working with trigonometric functions in SHM.
---4. Physical Realisations in Mechanics
In CCEA Mechanics 1, you will encounter three standard physical systems that exhibit Simple Harmonic Motion.
Model 1: Horizontal Elastic Spring or String System
Consider a particle of mass \(m\) on a smooth horizontal surface attached to an elastic string or spring of natural length \(l\) and modulus of elasticity \(\lambda\). The other end is fixed at a point \(A\).
Let the particle be displaced by a distance \(x\) from the natural length position \(O\). By Hooke's Law, the tension in the string is:
\(T = \frac{\lambda x}{l}\)
Applying Newton's Second Law (\(F = m\ddot{x}\)) in the direction of increasing \(x\):
\(m\ddot{x} = -T = -\frac{\lambda x}{l}\)
\(\ddot{x} = -\left(\frac{\lambda}{ml}\right)x\)
This matches the standard SHM form \(\ddot{x} = -\omega^2 x\) with:
\(\omega = \sqrt{\frac{\lambda}{ml}} \quad \text{and} \quad T = 2\pi\sqrt{\frac{ml}{\lambda}}\)
Model 2: Vertical Elastic Spring or String System
When a particle hangs vertically from an elastic spring or string, gravity stretches the spring by an equilibrium extension \(e\) before any oscillation begins.
Step 1: Equilibrium Position
At static equilibrium, the downward weight equals the upward tension:
\(mg = T_0 = \frac{\lambda e}{l}\)
Step 2: Displaced Position
Let \(x\) be the displacement measured downwards from the equilibrium position. The total extension is now \(e + x\).
Applying Newton's Second Law downwards:
\(m\ddot{x} = mg - T = mg - \frac{\lambda (e + x)}{l}\)
\(m\ddot{x} = mg - \frac{\lambda e}{l} - \frac{\lambda x}{l}\)
Since \(mg - \frac{\lambda e}{l} = 0\), the equation simplifies to:
\(m\ddot{x} = -\frac{\lambda x}{l} \implies \ddot{x} = -\left(\frac{\lambda}{ml}\right)x\)
Crucial Rule for Elastic Strings (The Slack String Condition):
An elastic string only exerts tension when it is stretched. If the amplitude \(a \le e\), the string remains taut throughout the motion, and the particle performs pure SHM.
If the amplitude \(a > e\), the particle rises above the natural length point. The string goes slack (\(T = 0\)), and the particle stops executing SHM, moving instead as a free projectile under gravity alone (\(\ddot{x} = -g\)) until it falls back to the natural length.
Model 3: The Simple Pendulum (Small Angle Approximation)
A simple pendulum consists of a mass \(m\) suspended from a light inextensible string of length \(l\). When displaced by an angle \(\theta\) from the vertical, the restoring force acting along the circular arc is the tangential component of weight:
\(m l \ddot{\theta} = -mg \sin\theta\)
\(\ddot{\theta} = -\frac{g}{l}\sin\theta\)
For small angular displacements (\(\theta\) is small, measured in radians), we use the small-angle approximation \(\sin\theta \approx \theta\):
\(\ddot{\theta} \approx -\left(\frac{g}{l}\right)\theta\)
This is SHM in angular form with:
\(\omega = \sqrt{\frac{g}{l}} \quad \text{and} \quad T = 2\pi\sqrt{\frac{l}{g}}\)
Worked Example 2: Vertical String System
Problem: A particle of mass \(0.5\text{ kg}\) is attached to the end of a light elastic string of natural length \(0.8\text{ m}\) and modulus of elasticity \(\lambda = 19.6\text{ N}\). The other end is attached to a fixed point. (Take \(g = 9.8\text{ m s}^{-2}\)).
(a) Find the extension of the string at equilibrium.
(b) Show that vertical oscillations about the equilibrium position are simple harmonic, and find the period.
(c) What is the maximum amplitude for which the motion is simple harmonic throughout?
Solution:
(a) At equilibrium, \(mg = \frac{\lambda e}{l}\):
\(0.5 \times 9.8 = \frac{19.6 e}{0.8}\)
\(4.9 = 24.5 e \implies e = \frac{4.9}{24.5} = 0.2\text{ m}\)
(b) Let \(x\) be the displacement downwards from equilibrium:
\(m\ddot{x} = mg - \frac{\lambda(e+x)}{l} = mg - \frac{\lambda e}{l} - \frac{\lambda x}{l}\)
Since \(mg = \frac{\lambda e}{l}\):
\(0.5\ddot{x} = -\frac{19.6}{0.8}x = -24.5x\)
\(\ddot{x} = -\frac{24.5}{0.5}x = -49x\)
Since this is in the form \(\ddot{x} = -\omega^2 x\) where \(\omega^2 = 49 \implies \omega = 7\text{ rad s}^{-1}\), the motion is SHM.
The period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{7}\text{ s} \approx 0.898\text{ s}\).
(c) The motion remains SHM as long as the string does not go slack. The string goes slack if the upward displacement exceeds the equilibrium extension \(e\). Therefore, the maximum amplitude is:
\(a_{\max} = e = 0.2\text{ m}\)
Key Takeaway: In vertical spring/string problems, displacement \(x\) must always be measured from the equilibrium position, not the unstretched position. For elastic strings, SHM only holds if amplitude \(a \le e\).
---5. Common Pitfalls and Top Exam Tips
• Degree Mode Trap: In SHM, \(\omega t\) is an angle in radians. If you calculate \(\cos(\omega t)\) or \(\sin(\omega t)\) with your calculator in degrees, your answer will be completely wrong. Always keep your calculator in radian mode.
• Measuring from the Wrong Origin: For vertical systems, always write the equation of motion relative to the static equilibrium point. Writing \(\ddot{x} = -\omega^2 x\) where \(x\) is measured from the suspension point or natural length point is mathematically incorrect.
• Ignoring Slack Strings: Remember that an elastic string cannot push. If a question states that an elastic string is pulled down by an amplitude greater than \(e\), the motion splits into two phases: SHM when taut (\(x \ge -e\)) and free vertical motion under gravity (\(\ddot{x} = -g\)) when slack.
• Finding Time Intervals Across the Centre: If asked for the time taken to travel between two points on opposite sides of the centre (e.g., from \(x = -1.5\text{ m}\) to \(x = +2.5\text{ m}\)), calculate the time from the centre to each point separately using \(x = a \sin(\omega t)\), and add the two times together.
---Quick Summary Reference Sheet
• Governing Equation: \(\ddot{x} = -\omega^2 x\)
• Velocity–Displacement: \(v^2 = \omega^2 (a^2 - x^2)\)
• Periodic Time: \(T = \frac{2\pi}{\omega}\)
• Frequency: \(f = \frac{\omega}{2\pi}\)
• Maximum Speed: \(v_{\max} = a\omega\) (at centre \(x = 0\))
• Maximum Acceleration: \(a_{\max} = a\omega^2\) (at extremes \(x = \pm a\))
• Displacement from centre (\(t=0, x=0\)): \(x = a\sin(\omega t)\)
• Displacement from extreme (\(t=0, x=a\)): \(x = a\cos(\omega t)\)
• Horizontal/Vertical Spring System: \(\omega = \sqrt{\frac{\lambda}{ml}}\), \(T = 2\pi\sqrt{\frac{ml}{\lambda}}\)
• Simple Pendulum (small angles): \(\omega = \sqrt{\frac{g}{l}}\), \(T = 2\pi\sqrt{\frac{l}{g}}\)