Simple Harmonic Motion Equations, Mass-Spring & Pendulum Systems, Damping & Resonance
Simple Harmonic Motion (SHM) is the universal mathematical model for oscillatory systems, from a mass on a helical spring and a simple pendulum to vibrating quartz crystals and diatomic covalent bonds. An oscillation is strictly defined as SHM if and only if the acceleration a is directly proportional to the displacement x from the equilibrium position and is always directed toward the equilibrium position: a = -ω²x, where ω = 2πf = 2π/T is the angular frequency. Because velocity is the time derivative of displacement and acceleration is the time derivative of velocity, the x-t, v-t, and a-t sinusoidal graphs are shifted successively by π/2 radians (90°) in phase, while kinetic and potential energy exchange continuously so total mechanical energy E_total = ½mω²A² remains constant. Examiners test defining SHM from graphs, fiducial marker timing techniques, light/critical/heavy damping, and forced resonance curves.
- Defining Equation of Simple Harmonic Motion: a = -ω²x
- Two conditions define Simple Harmonic Motion (SHM): 1. Acceleration a is directly proportional to displacement x from the equilibrium position (a ∝ x). 2. Acceleration (and restoring force F = -kx) is always directed toward the equilibrium position (opposite in direction to displacement, shown by the minus sign in a = -ω²x). Here ω = 2πf = 2π / T is angular frequency (in rad s⁻¹). A graph of Acceleration a (y-axis) against Displacement x (x-axis) is always a straight line through the origin with a negative gradient equal to -ω².
- Displacement, Velocity & Acceleration Equations & Phase Graphs
- If timing starts (t = 0) when the oscillator is released from maximum positive displacement (+A, amplitude): 1. Displacement: x = A cos(ωt) (or x = A sin(ωt) if starting at equilibrium x = 0 at t = 0). 2. Velocity: v = -Aω sin(ωt) = ±ω√(A² - x²); maximum speed v_max = ωA occurs at the equilibrium position (x = 0), and v = 0 at the extremes (x = ±A). 3. Acceleration: a = -Aω² cos(ωt) = -ω²x; maximum magnitude of acceleration a_max = ω²A occurs at the extremes (x = ±A), and a = 0 at equilibrium (x = 0). Phase relationships: v leads x by π/2 rad (90°); a leads v by π/2 rad and is in antiphase (π rad / 180°) with x.
- Mass-Spring System (T = 2π√(m / k)) vs Simple Pendulum (T = 2π√(L / g))
- 1. Mass-Spring System: time period T = 2π√(m / k), where m is mass (in kg) and k is spring constant (in N m⁻¹). Notice that T is independent of amplitude A (isochronous) and independent of gravitational field strength g (a mass-spring system has the exact same time period on the Moon or in an orbiting space station as on Earth).To linearise: plot T² against m (straight line through origin with gradient = 4π² / k). 2. Simple Pendulum (valid for small angles θ < 10° where sin θ ≈ θ in radians): time period T = 2π√(L / g), where L is length from pivot to centre of mass of the bob. Notice T is independent of the mass m of the bob. To linearise to find g: plot T² against L (straight line through origin with gradient = 4π² / g, so g = 4π² / gradient).
Simple Harmonic Motion Equations, Mass-Spring & Pendulum Systems, Damping & Resonance: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
v = f × λ = 4 Hz × 2.5 m = 10.0 m/sTime Period: T = 1 ÷ 4 = 0.25 sKey Concepts & Exam Definitions (6)
- Defining Equation of Simple Harmonic Motion: a = -ω²x
- Two conditions define Simple Harmonic Motion (SHM): 1. Acceleration a is directly proportional to displacement x from the equilibrium position (a ∝ x). 2. Acceleration (and restoring force F = -kx) is always directed toward the equilibrium position (opposite in direction to displacement, shown by the minus sign in a = -ω²x). Here ω = 2πf = 2π / T is angular frequency (in rad s⁻¹). A graph of Acceleration a (y-axis) against Displacement x (x-axis) is always a straight line through the origin with a negative gradient equal to -ω².
- Displacement, Velocity & Acceleration Equations & Phase Graphs
- If timing starts (t = 0) when the oscillator is released from maximum positive displacement (+A, amplitude): 1. Displacement: x = A cos(ωt) (or x = A sin(ωt) if starting at equilibrium x = 0 at t = 0). 2. Velocity: v = -Aω sin(ωt) = ±ω√(A² - x²); maximum speed v_max = ωA occurs at the equilibrium position (x = 0), and v = 0 at the extremes (x = ±A). 3. Acceleration: a = -Aω² cos(ωt) = -ω²x; maximum magnitude of acceleration a_max = ω²A occurs at the extremes (x = ±A), and a = 0 at equilibrium (x = 0). Phase relationships: v leads x by π/2 rad (90°); a leads v by π/2 rad and is in antiphase (π rad / 180°) with x.
- Mass-Spring System (T = 2π√(m / k)) vs Simple Pendulum (T = 2π√(L / g))
- 1. Mass-Spring System: time period T = 2π√(m / k), where m is mass (in kg) and k is spring constant (in N m⁻¹). Notice that T is independent of amplitude A (isochronous) and independent of gravitational field strength g (a mass-spring system has the exact same time period on the Moon or in an orbiting space station as on Earth).To linearise: plot T² against m (straight line through origin with gradient = 4π² / k). 2. Simple Pendulum (valid for small angles θ < 10° where sin θ ≈ θ in radians): time period T = 2π√(L / g), where L is length from pivot to centre of mass of the bob. Notice T is independent of the mass m of the bob. To linearise to find g: plot T² against L (straight line through origin with gradient = 4π² / g, so g = 4π² / gradient).
- Energy in Undamped SHM (E_k, E_p, and Total Energy E_total)
- In an undamped oscillator, mechanical energy continuously interconverts between kinetic energy (E_k = ½mv² = ½mω²(A² - x²), inverted parabola peaking at x = 0) and potential energy (E_p = ½kx² = ½mω²x², upward parabola peaking at x = ±A). The total mechanical energy is constant at all points and proportional to amplitude squared: E_total = E_k + E_p = ½mω²A² = ½kA². Note: Both E_k and E_p complete two full cycles during one time period T of the oscillation.
- Free vs Forced Oscillations & Damping (Light, Critical, Heavy)
- Free oscillations occur at the system's natural frequency f₀ with no external periodic driving force. Damping is the removal of mechanical energy from an oscillating system by dissipative resistive forces (air resistance, viscous drag, friction), converting it into thermal energy: 1. Light damping (underdamping): amplitude decays exponentially over many oscillations while time period T stays virtually constant. 2. Critical damping: reduces displacement to equilibrium (x = 0) in the shortest possible time (¼T) without overshooting or oscillating (used in car suspension shock absorbers and analogue meter pointers). 3. Heavy damping (overdamping): returns very slowly to equilibrium without oscillating.
- Resonance & Effect of Damping on Resonance Curves
- Forced oscillations occur when a periodic external driving force of frequency f_d is applied. Resonance occurs when the driving frequency equals the natural frequency of the system (f_d ≈ f₀): maximum energy is transferred from the driver to the oscillator per cycle, causing the amplitude of oscillation to reach a sharp maximum (with a π/2 rad phase lag between driver and oscillator). Increasing the degree of damping on a resonance curve (Amplitude vs Driving Frequency): 1. Lowers the peak amplitude at resonance, 2. Broadens (flattens) the resonance peak, and 3. Shifts the resonant peak frequency slightly to the left (slightly below f₀). Examples: MRI proton spin resonance, microwave ovens, radio tuning circuits, vs destructive wind-induced resonance on bridges (Tacoma Narrows / Millennium Bridge) and seismic building dampers.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the two conditions that define Simple Harmonic Motion (SHM) and describe the shape of an acceleration-displacement (a vs x) graph.
Where in an SHM cycle do maximum speed (v_max) and maximum acceleration (a_max) occur, and what are their formulas?
Why should the fiducial marker (pin) be placed at the centre of oscillation (equilibrium position) rather than at the extreme amplitude when timing 20 oscillations with a stopwatch?
If a simple pendulum and a vertical mass-spring system are both taken from Earth (g = 9.81 m s⁻²) to the Moon (g = 1.62 m s⁻²), how does the time period T of each change?
What is critical damping, and give a real-world engineering application where it is needed?
State three changes to the Amplitude vs Driving Frequency graph when damping is increased in a driven oscillating system.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A piston of mass 0.450 kg in a car engine oscillates with approximate simple harmonic motion of amplitude 42.0 mm at a frequency of 50.0 Hz. Calculate: (a) the angular frequency ω, (b) the maximum speed of the piston, (c) the magnitude of the maximum resultant force acting on the piston, and (d) the speed of the piston when it is 25.0 mm from the equilibrium position. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Angular frequency ω = 2πf = 2π × 50.0 = 314.2 rad s⁻¹ (100π rad s⁻¹) [1 mark]. 2. (b) Maximum speed (at x = 0): v_max = ωA = 314.2 rad s⁻¹ × 0.0420 m = 13.2 m s⁻¹ [1 mark]. 3. (c) Maximum acceleration (at x = ±A): a_max = ω²A = (314.2)² × 0.0420 = 4.145 × 10³ m s⁻² [1 mark]. 4. Maximum resultant force F_max = m a_max = 0.450 kg × 4.145 × 10³ m s⁻² = 1.87 × 10³ N (1865 N) [1 mark]. 5. (d) Speed at x = 0.0250 m using v = ω√(A² - x²): v = 314.2 × √((0.0420)² - (0.0250)²) = 314.2 × √(0.001764 - 0.000625) = 314.2 × 0.03375 = 10.6 m s⁻¹ [1 mark].
- Q2
A student investigated how the time period T of a simple pendulum varies with its length L to determine the acceleration due to gravity g. Describe the experimental procedure they should follow to obtain an accurate value for g using a linear graph, including how T² is related to L. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Measure the length L from the point of suspension (clamp) to the centre of mass of the pendulum bob using a metre rule, and place a vertical fiducial marker (pin) behind the string at the equilibrium position [1 mark]. 2. Displace the bob by a small angle (θ < 10° so the motion is SHM) and use a stopwatch to measure the time for 20 complete oscillations (20T), starting and stopping at the fiducial marker; repeat 3 times, calculate the mean, and divide by 20 to find T [1 mark]. 3. Repeat for at least 6 different lengths L (e.g. from 0.30 m to 1.20 m) [1 mark]. 4. Square the equation T = 2π√(L / g) to give T² = (4π² / g)L (of the form y = mx) and plot a graph of T² on the y-axis against L on the x-axis [1 mark]. 5. Draw a line of best fit through the origin, measure the gradient using a large triangle, and calculate g = 4π² ÷ gradient [1 mark].
- Q3
A trolley of mass 0.800 kg is tethered horizontally between two springs with a combined effective spring constant k = 72.0 N m⁻¹ on a frictionless air track. The trolley is pulled 0.150 m to the right of equilibrium and released from rest. Calculate the time period of oscillation, the total energy of the oscillating system, and the kinetic and potential energies when the displacement is 0.100 m. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Time period T = 2π√(m ÷ k) = 2π√(0.800 ÷ 72.0) = 2π√(1/90) = 0.662 s [1 mark]. 2. Total energy equals the maximum elastic potential energy at x = A = 0.150 m: E_total = ½kA² = 0.5 × 72.0 × (0.150)² = 0.810 J [1 mark]. 3. At x = 0.100 m, potential energy E_p = ½kx² = 0.5 × 72.0 × (0.100)² = 0.360 J [1 mark]. 4. By conservation of energy, kinetic energy at x = 0.100 m is E_k = E_total - E_p = 0.810 - 0.360 = 0.450 J [1 mark].
- Q4
Explain what is meant by resonance, describe why resonance occurred when pedestrians walked across the London Millennium Footbridge, and explain how installing viscous fluid dampers solved the problem. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Resonance occurs when the frequency of an external periodic driving force equals the natural frequency of an oscillating system, resulting in maximum energy transfer and a very large amplitude of oscillation [1 mark]. 2. The footsteps of pedestrians walking across the bridge provided a periodic lateral driving force whose frequency matched the natural lateral oscillation frequency of the bridge structure [1 mark]. 3. As the bridge began to sway with a larger amplitude, pedestrians instinctively adjusted their step timing to stay balanced, synchronising their footsteps in phase with the bridge and feeding even more energy into the resonant oscillation [1 mark]. 4. Viscous fluid dampers absorb kinetic energy from the oscillating bridge and dissipate it as thermal energy in the fluid during every cycle, drastically reducing the peak amplitude at the resonant frequency [1 mark].
Common Exam Questions (4)
Why must my calculator be in radians mode when substituting a time t into x = A cos(ωt) or v = -Aω sin(ωt)?
Because angular frequency ω = 2πf is measured in radians per second (rad s⁻¹), the angle (ωt) inside the cosine or sine brackets is always in radians. If your calculator is left in degrees mode, x = A cos(ωt) will give a completely wrong answer.
If the amplitude A of an SHM oscillator is doubled, how do its time period T, maximum speed v_max, maximum acceleration a_max, and total energy E_total change?
1. Time period T (and frequency f and ω): Unchanged (SHM is isochronous; T = 2π√(m/k) does not contain A). 2. Maximum speed v_max = ωA: Doubles (×2). 3. Maximum acceleration a_max = ω²A: Doubles (×2). 4. Total energy E_total = ½mω²A²: QUADRUPLES (×4)!
Why is the simple pendulum only an approximate SHM oscillator for small angles (θ < 10°)?
The restoring force tangential to the arc of a pendulum bob is F = -mg sin θ, so a = -g sin θ. Only for small angles (θ < 10° or ~0.17 rad) does the small-angle approximation sin θ ≈ θ = x / L hold true, giving a ≈ -(g / L)x (where ω² = g / L).
At what fraction of the amplitude A is the kinetic energy equal to the potential energy in SHM?
Set E_p = ½ E_total: ½mω²x² = ½(½mω²A²) → x² = ½A² → x = ±A / √2 ≈ ±0.707 A (not at half the amplitude x = 0.5 A, where E_p is only ¼ of E_total and E_k is ¾ of E_total).
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