Representing & Analyzing
Simple Harmonic Motion
Master the equations, graphs, and energy analysis of oscillatory systems: from mass-spring setups to pendulums.
AP Physics 1
What is Simple Harmonic Motion?
Simple Harmonic Motion (SHM) is a type of periodic oscillation where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction.
Periodic
The motion repeats at regular time intervals called the period \( T \).
Restoring
The force always points toward equilibrium, opposing displacement.
Sinusoidal
Position, velocity, and acceleration follow sine and cosine functions.
Key Quantities in SHM
Amplitude \( A \)
Maximum displacement from equilibrium. Measured in meters.
Period \( T \)
Time for one complete oscillation. Measured in seconds.
Frequency \( f \)
Oscillations per second: \( f = \frac{1}{T} \). Measured in Hz.
Angular Frequency \( \omega \)
\( \omega = 2\pi f = \frac{2\pi}{T} \). Measured in rad/s.
Visualizing SHM: \( x(t) \) Graph
The Core Equations of SHM
1
Position: \( x(t) = A\cos(\omega t + \phi) \)
2
Velocity: \( v(t) = -A\omega\sin(\omega t + \phi) \)
3
Acceleration: \( a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x \)
4
Maximum velocity: \( v_{\max} = A\omega \)
5
Maximum acceleration: \( a_{\max} = A\omega^2 \)
Period of a Mass-Spring System
Period of a Simple Pendulum
Mass-Spring vs Pendulum
| Mass-Spring | Pendulum | |
|---|---|---|
| Period formula | \( T = 2\pi\sqrt{m/k} \) | \( T = 2\pi\sqrt{L/g} \) |
| Depends on mass? | Yes | No |
| Depends on amplitude? | No | No (small angles) |
| Restoring force | \( F = -kx \) | \( F \approx -\frac{mg}{L}x \) |
| Valid when | Ideal spring (Hooke's law) | Small angle (\( \theta < 15^\circ \)) |
Energy in Simple Harmonic Motion
Velocity at Any Position
From energy conservation, we can derive the velocity at any displacement \( x \):
At \( x = 0 \)
\( v = \pm A\omega \): maximum speed
At \( x = \pm A \)
\( v = 0 \): momentarily at rest
How to Analyze SHM Problems
1
Identify the system: Is it a mass-spring or a pendulum? This determines which period formula to use.
2
Find \( \omega \): Calculate angular frequency from \( T \) or \( f \) using \( \omega = 2\pi f = 2\pi / T \).
3
Determine amplitude: Identify \( A \) from initial conditions (max displacement, or energy).
4
Write the equations: Use \( x(t) = A\cos(\omega t + \phi) \) and find \( \phi \) from initial position/velocity.
5
Apply energy methods: Use \( E = \frac{1}{2}kA^2 \) when the problem involves speed at a given position.
Example 1: Finding Period and Frequency
A 0.50 kg block is attached to a spring with \( k = 200 \) N/m. Find the period, frequency, and angular frequency.
1
Period: \( T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.50}{200}} = 2\pi\sqrt{0.0025} = 2\pi(0.05) \approx 0.314 \text{ s} \)
2
Frequency: \( f = \frac{1}{T} = \frac{1}{0.314} \approx 3.18 \text{ Hz} \)
3
Angular frequency: \( \omega = 2\pi f = 2\pi(3.18) \approx 20.0 \text{ rad/s} \)
Example 2: Speed at a Given Position
The same block (\( m = 0.50 \text{ kg} \), \( k = 200 \text{ N/m} \)) is pulled 0.10 m from equilibrium and released. What is its speed when \( x = 0.06 \text{ m} \)?
1
Identify: \( A = 0.10 \text{ m} \), \( \omega = 20.0 \text{ rad/s} \), \( x = 0.06 \text{ m} \)
2
Apply velocity formula: \( v = \omega\sqrt{A^2 - x^2} \)
3
Calculate: \( v = 20.0\sqrt{(0.10)^2 - (0.06)^2} = 20.0\sqrt{0.0064} = 20.0 \times 0.08 = 1.6 \text{ m/s} \)
Example 3: Writing the Position Equation
A pendulum of length \( L = 1.0 \text{ m} \) swings with a small amplitude of 0.05 m. At \( t = 0 \), the bob is at \( x = 0 \) and moving in the positive direction. Write \( x(t) \).
1
Find \( \omega \): \( \omega = \sqrt{g/L} = \sqrt{9.8/1.0} \approx 3.13 \text{ rad/s} \)
2
At \( t = 0 \): \( x(0) = 0 \) and \( v(0) > 0 \). Using \( x = A\cos(\omega t + \phi) \), we need \( \cos(\phi) = 0 \) and \( -\sin(\phi) > 0 \).
3
Phase: \( \phi = -\frac{\pi}{2} \), so \( x(t) = A\cos(\omega t - \frac{\pi}{2}) = A\sin(\omega t) \)
Graphs of \( x \), \( v \), and \( a \)
A rotating peg whose shadow becomes an oscillator
One coordinate of a peg in uniform circular motion varies sinusoidally. The spring cart provides a second oscillator model in which acceleration points toward equilibrium.A rotating peg whose shadow becomes an oscillator
One coordinate of a peg in uniform circular motion varies sinusoidally. The spring cart provides a second oscillator model in which acceleration points toward equilibrium.Common Mistakes in SHM
✕ Amplitude affects period
For ideal SHM, the period is independent of amplitude. Doubling \( A \) does not change \( T \).
✕ Confusing \( v_{\max} \) and \( a_{\max} \)
\( v_{\max} = A\omega \) at equilibrium. \( a_{\max} = A\omega^2 \) at the extremes. They occur at different positions.
✕ Wrong phase constant
If the object starts at equilibrium, use \( \phi = \pm\pi/2 \), not \( \phi = 0 \). Check both \( x(0) \) and \( v(0) \) direction.
Summary of Key SHM Formulas
| Quantity | Formula | Units |
|---|---|---|
| Position | \( x = A\cos(\omega t + \phi) \) | m |
| Velocity | \( v = -A\omega\sin(\omega t + \phi) \) | m/s |
| Acceleration | \( a = -\omega^2 x \) | \( \text{m/s}^2 \) |
| Period (spring) | \( T = 2\pi\sqrt{m/k} \) | s |
| Period (pendulum) | \( T = 2\pi\sqrt{L/g} \) | s |
| Total energy | \( E = \frac{1}{2}kA^2 \) | J |
| Speed at position | \( v = \omega\sqrt{A^2 - x^2} \) | m/s |
Question 1 of 5
A mass on a spring oscillates with \( A = 0.20 \text{ m} \) and \( T = 0.50 \text{ s} \). What is the maximum speed of the mass?
\( v_{\max} = A\omega = A \cdot \frac{2\pi}{T} = 0.20 \cdot \frac{2\pi}{0.50} = 0.20 \times 12.57 \approx 2.51 \text{ m/s} \). Choice (b) uses \( 2\pi A \) without dividing by \( T \); choice (a) uses \( A/T \).
Question 2 of 5
If you double the mass on a spring (keeping \( k \) constant), the period will:
Since \( T = 2\pi\sqrt{m/k} \), doubling \( m \) gives \( T' = 2\pi\sqrt{2m/k} = \sqrt{2} \cdot T \). The period increases by \( \sqrt{2} \approx 1.41 \), not by 2.
Question 3 of 5
In SHM, at what displacement does the kinetic energy equal the potential energy?
When \( KE = PE \), each equals \( \frac{1}{2}E_{\text{total}} \). Setting \( \frac{1}{2}kx^2 = \frac{1}{4}kA^2 \) gives \( x^2 = A^2/2 \), so \( x = A/\sqrt{2} \approx 0.707A \). At \( x = A/2 \), PE is only \( \frac{1}{4}E \).
Question 4 of 5
A simple pendulum has \( T = 2.0 \text{ s} \) on Earth. If taken to a planet where \( g = 4.9 \text{ m/s}^2 \), its new period is:
\( T = 2\pi\sqrt{L/g} \). Halving \( g \) (from 9.8 to 4.9) multiplies \( T \) by \( \sqrt{2} \): \( T' = 2.0 \times \sqrt{2} \approx 2.83 \text{ s} \). Gravity affects the pendulum period, unlike a spring.
Question 5 of 5
A block on a spring has \( A = 0.12 \text{ m} \) and \( \omega = 10 \text{ rad/s} \). What is its maximum acceleration?
\( a_{\max} = A\omega^2 = 0.12 \times (10)^2 = 0.12 \times 100 = 12 \text{ m/s}^2 \). Choice (a) uses \( A\omega \) (which gives \( v_{\max} \)); choice (c) uses \( A\omega^3 \).