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Simple Harmonic
Motion

Explore the physics of oscillations, understand the restoring force, and master the equations that describe periodic motion from springs to pendulums.

AP Physics 1

What Is Oscillation?

An oscillation is any repetitive back-and-forth motion around a central point called the equilibrium position. When you pull a mass on a spring and release it, the mass moves back and forth: that is oscillation.

Equilibrium
The position where the net force is zero and the object naturally rests
Periodic
The motion repeats itself at regular time intervals called the period
Restoring Force
A force that always pushes the object back toward equilibrium

Defining Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a special type of oscillation where the restoring force is directly proportional to the displacement from equilibrium and always directed toward it.

This linear relationship between force and displacement is what makes the motion "simple." If the restoring force has any other dependence on \( x \), the motion is oscillatory but not simple harmonic.

Visualizing SHM: Mass-Spring System

Equilibrium m F
\(x < 0\)
 m F
\(x > 0\)

A numerical illustration of the restoring force

This separate spring example illustrates the same force directions with m = 0.50 kg, k = 200 N/m, and amplitude 0.10 m. Its period and maximum speed apply to these values.

A spring exerts a restoring force. For m = 0.50 kg and k = 200 N/m, displacement −0.10 m gives rightward force, zero displacement gives zero force, and +0.10 m gives leftward force. T = 0.314 s and maximum speed is 2 m/s.
A spring exerts a restoring force. For m = 0.50 kg and k = 200 N/m, displacement −0.10 m gives rightward force, zero displacement gives zero force, and +0.10 m gives leftward force. T = 0.314 s and maximum speed is 2 m/s.

The Equation of Motion

Applying Newton's second law \( F = ma \) to the restoring force \( F = -kx \) gives us the fundamental differential equation of SHM:

The acceleration is always proportional to and opposite in direction to the displacement. This is the hallmark of SHM: \( a \propto -x \).

Position, Velocity, and Acceleration

The solution to \( a = -\omega^2 x \) gives position as a sinusoidal function of time. Velocity and acceleration follow by differentiation:

Position
\( x(t) = A\cos(\omega t + \phi) \)
Velocity
\( v(t) = -A\omega\sin(\omega t + \phi) \)
Acceleration
\( a(t) = -A\omega^2\cos(\omega t + \phi) \)

A rotating peg whose shadow becomes an oscillator

Bench with lamp, peg wheel, blank screen and a spring connecting a fixed support to a sliding cart.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

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.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.

Graphing SHM Over Time

t + −
\(T/2\)
 T
\(x(t)\)
\(v(t)\)
\(a(t)\)

Key SHM Parameters

1
Amplitude \( A \): Maximum displacement from equilibrium. Always positive.
2
Period \( T \): Time for one complete oscillation. \( T = \frac{2\pi}{\omega} \).
3
Frequency \( f \): Number of oscillations per second. \( f = \frac{1}{T} = \frac{\omega}{2\pi} \).
4
Angular Frequency \( \omega \): Rate of oscillation in rad/s. \( \omega = 2\pi f = \sqrt{k/m} \) for a spring.
5
Phase Constant \( \phi \): Determines the initial position and velocity at \( t = 0 \).

SHM vs. General Oscillation

Property Simple Harmonic General Oscillation
Restoring Force \( F = -kx \) (linear) Any function of \( x \)
Motion Shape Perfectly sinusoidal May be non-sinusoidal
Period Independent of amplitude May depend on amplitude
Examples Ideal spring, small-angle pendulum Large-angle pendulum, bouncing ball

Energy in SHM

In SHM, energy continuously converts between kinetic and potential forms while the total mechanical energy remains constant (no friction).

Kinetic Energy
\( KE = \frac{1}{2}mv^2 \)
Maximum at equilibrium, zero at endpoints
Potential Energy
\( PE = \frac{1}{2}kx^2 \)
Maximum at endpoints, zero at equilibrium

Energy Diagram in SHM

x E −A +A E total PE KE

Period of a Mass-Spring System

Increase \( m \)
Heavier mass oscillates more slowly: period increases
Increase \( k \)
Stiffer spring oscillates faster: period decreases

Period of a Simple Pendulum

\(m\)
\(\theta\)
\(L\)
\(mg\)

How to Identify SHM

1
Find the equilibrium position: Where does the net force on the object equal zero?
2
Displace from equilibrium: If displaced by a small amount \( x \), does a restoring force appear?
3
Check proportionality: Is the restoring force proportional to \( x \)? Does \( F = -kx \) hold?
4
Confirm linearity: If \( F \propto -x \), the motion is SHM. If the relationship is nonlinear, it is not SHM.

Example 1: Spring Period

A \(0.50 \text{ kg}\) block is attached to a spring with spring constant \( k = 200 \text{ N/m}\). Find the period and frequency of oscillation.

1
Identify the formula: \( T = 2\pi\sqrt{m/k} \)
2
Substitute values: \( T = 2\pi\sqrt{0.50 \text{ kg} / 200 \text{ N/m}} = 2\pi\sqrt{0.0025 \text{ s}^2} \)
3
Calculate: \( T = 2\pi(0.05 \text{ s}) \approx 0.314 \text{ s}\)
4
Find frequency: \( f = 1/T = 1/(0.314 \text{ s}) \approx 3.18 \text{ Hz}\)

Example 2: Maximum Speed

The same block (\( m = 0.50 \text{ kg} \), \( k = 200 \text{ N/m} \)) is pulled \(0.08 \text{ m}\) from equilibrium and released. Find the maximum speed.

1
Use energy conservation: At equilibrium, all PE becomes KE: \( \frac{1}{2}kA^2 = \frac{1}{2}mv_{\max}^2 \)
2
Solve for \( v_{\max} \): \( v_{\max} = A\sqrt{k/m} = A\omega \)
3
Substitute: \( v_{\max} = (0.08 \text{ m}) \times \sqrt{(200 \text{ N/m})/(0.50 \text{ kg})} \approx 1.6 \text{ m/s}\)

Common Mistakes

✕ Period Depends on Amplitude
In SHM, the period is independent of amplitude. A larger amplitude does not make it oscillate slower or faster.
✕ Confusing \( \omega \) and \( f \)
Angular frequency \( \omega \) is in rad/s, while frequency \( f \) is in Hz. They differ by a factor of \( 2\pi \): \( \omega = 2\pi f \).
✕ Forgetting the Small-Angle Approximation
A pendulum is SHM only for small angles. The formula \( T = 2\pi\sqrt{L/g} \) fails for large swings where \( \sin\theta \not\approx \theta \).

SHM Summary

Quantity Formula Units
Position \( x(t) = A\cos(\omega t + \phi) \) \(\text{m}\)
Velocity \( v(t) = -A\omega\sin(\omega t + \phi) \) \(\text{m/s}\)
Acceleration \( a(t) = -\omega^2 x \) \(\text{m/s}^2\)
Period (spring) \( T = 2\pi\sqrt{m/k} \) \(\text{s}\)
Period (pendulum) \( T = 2\pi\sqrt{L/g} \) \(\text{s}\)
Max Speed \( v_{\max} = A\omega \) \(\text{m/s}\)
Total Energy \( E = \frac{1}{2}kA^2 \) \(\text{J}\)

Question 1 of 5

Which condition must be true for a system to undergo simple harmonic motion?

Question 2 of 5

A mass on a spring has period \( T \). If the mass is quadrupled, what is the new period?

Question 3 of 5

In SHM, at what position does the object have maximum speed?

Question 4 of 5

A simple pendulum has length \( L \) and period \( T \) on Earth. If taken to a planet where \( g \) is four times Earth's gravity, what is the new period?

Question 5 of 5

If the amplitude of a mass-spring SHM system is doubled, what happens to the total mechanical energy?