8.1 Simple harmonic motion

Achievable AP Physics 1

8. Oscillations

Our AP Physics 1 course is currently in development and is a work-in-progress.

Simple harmonic motion

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When an object oscillates back and forth about an equilibrium position under a restoring force proportional to its displacement, the motion is called simple harmonic motion (SHM). Here are some real-life scenarios that show the same basic idea:

A car’s suspension is a tuned bounce-and-return system that helps keep the wheels on the road.
Skyscrapers can sway in the wind; to reduce the motion, builders sometimes install large pendulums near the top - heavy swinging masses that counter the sway.
A phone’s vibration motor spins an off-center weight, which makes the phone’s casing oscillate in a quick, controlled way.

In SHM, the acceleration always points toward equilibrium and satisfies

$a = - ω^{2} x,$

where

$ω \equiv \frac{k}{m} (rad/s)$

is the angular frequency, $m$ is the mass, $k$ is the spring constant, and $x$ is the displacement from equilibrium.

Mass-spring derivation

Free‐body diagram:

Spring mass system in oscillatory motion

Equation of motion. Apply Newton’s second law along the horizontal direction:

$\sum F_{x} = m a_{x}, - k x = m \overset{x}{¨} .$

Rearrange:

$\overset{x}{¨} + \frac{k}{m} x = 0.$

Define

$ω \equiv \frac{k}{m},$

so the differential equation becomes

$\overset{x}{¨} + ω^{2} x = 0.$

You don’t need to do the calculus to use SHM results in most problems. The derivatives here are included to show how the standard formulas fit together.

A standard solution to this equation is

$x (t) = A cos (ω t + ϕ),$

where $A$ is the amplitude (maximum displacement) and $ϕ$ is the phase constant (set by the initial conditions). Once $x (t)$ is known, velocity and acceleration come from differentiation:

$v (t) = \overset{x}{˙} (t) = - A ω sin (ω t + ϕ),$

$a (t) = \overset{x}{¨} (t) = - A ω^{2} cos (ω t + ϕ) = - ω^{2} x (t) .$

Thus:

Velocity leads displacement by a quarter-cycle (90°).
Acceleration is in phase with displacement but opposite in sign.

Period and frequency in SHM

In simple harmonic motion, the period $T$ is the time required to complete one full oscillation (returning to the same displacement and velocity). The frequency $f$ is the number of complete oscillations per unit time. By definition,

$f = \frac{1}{T} .$

Since the displacement in SHM can be written

$x (t) = A cos (ω t + ϕ),$

the argument $ω t$ must increase by $2 π$ for one full cycle. That gives

$ω T = 2 π ⟹ T = \frac{2 π}{ω} .$

Because $ω = k / m$ for a mass-spring system (or $g / L$ for a small-angle pendulum), the period depends only on the system’s parameters:

$T_{spring} = 2 π \frac{m}{k}, T_{pendulum} = 2 π \frac{L}{g} .$

All SHM systems with the same $ω$ share the same $T$ and $f$ , regardless of amplitude $A$ .
Higher $ω$ (stronger restoring force or smaller inertia) $\Rightarrow$ shorter $T$ $\Rightarrow$ higher $f$ .
Frequency is measured in hertz (Hz), where $1 Hz = 1$ cycle/sec.

Graphical representation of SHM

Normalized SHM plots for position, velocity, and acceleration over one full cycle.

Here the solid blue curve is $x / A = cos (ω t + ϕ)$ and the red dashed curve is $v / (A ω) = - sin (ω t + ϕ)$ .

Simple pendulum (small‐angle approximation)

A simple pendulum consists of a point mass $m$ suspended from a fixed pivot by a light rod or string of length $L$ . If you displace it by a small angle and release it, it executes simple harmonic motion under gravity.

Restoring force and equation of motion

At an angular displacement $θ$ , the component of the weight $m g$ acting along the arc is

$F_{tan} = - m g sin θ .$

For small angles $(θ ≪ 1)$ , $sin θ \approx θ$ . If $s$ is the arc length from equilibrium, then

$s = L θ ⟹ sin θ \approx θ = \frac{s}{L},$

so the tangential force becomes

$F_{tan} \approx - m g \frac{s}{L} .$

Apply Newton’s second law in the tangential direction:

$m \overset{s}{¨} = - m g \frac{s}{L} ⟹ \overset{s}{¨} + \frac{g}{L} s = 0.$

This matches the SHM form $\overset{s}{¨} + ω^{2} s = 0$ with

$ω = \frac{g}{L} (rad/s) .$

General solution

The displacement along the arc varies sinusoidally:

$s (t) = S cos (ω t + ϕ),$

where $S$ is the maximum arc displacement (amplitude) and $ϕ$ is the phase constant set by initial conditions. Differentiating,

$v (t) = \overset{s}{˙} = - S ω sin (ω t + ϕ), a (t) = \overset{s}{¨} = - ω^{2} S cos (ω t + ϕ) = - ω^{2} s (t) .$

Period and frequency

One full oscillation corresponds to the argument of the cosine increasing by $2 π$ :

$ω T = 2 π ⟹ T = \frac{2 π}{ω} = 2 π \frac{L}{g}, f = \frac{1}{T} .$

In the small-angle limit, the pendulum behaves like a mass-spring system, with $ω = g / L$ .

Example Problem 1

A block of mass $m = 0.200$ kg is attached to a horizontal spring of constant $k = 50.0$ N/m on a frictionless table. The block is pulled to the right and released from rest at a displacement $x = A = 0.100$ m. (a) Find the angular frequency $ω$ and period $T$ .

$ω = \frac{k}{m} = \frac{50.0}{0.200} = 250 = 15.81 rad/s, T = \frac{2 π}{ω} = \frac{2 π}{15.81} = 0.397 s .$

(b) Write the equation of motion $x (t)$ assuming $t = 0$ at release.

Released from rest at $x = A$ implies phase $ϕ = 0$ , so

$x (t) = A cos (ω t) = 0.100 cos (15.81 t) m .$

(c) Determine the block’s speed when it passes through $x = 0.060$ m.

In SHM,

$v^{2} = ω^{2} (A^{2} - x^{2}) .$

Thus

$v = \pm ω A^{2} - x^{2} = 15.81 0.0064 = 1.265 m/s .$

Since the block is moving toward equilibrium, take $v = + 1.27$ m/s.

(d) Determine the acceleration of the block at $x = 0.060$ m.

Acceleration in SHM is $a = - ω^{2} x$ :

$a = - (15.81)^{2} (0.060) = - 250 (0.060) = - 15.0 m/ s^{2} .$

The negative sign shows the acceleration is toward equilibrium (leftward).

Example Problem 2

A simple pendulum of length $L = 1.00$ m oscillates with a maximum angular displacement $θ_{m a x} = 0.100$ rad (about $5.7 3^{\circ}$ ). Take $g = 9.80$ m/s $^{2}$ . (a) Compute the angular frequency $ω$ and period $T$ .

For small angles,

$ω = \frac{g}{L} = \frac{9.80}{1.00} = 3.130 rad/s, T = \frac{2 π}{ω} = \frac{2 π}{3.130} = 2.01 s .$

(b) Write $θ (t)$ assuming $θ (0) = Θ = θ_{m a x}$ and released from rest.

Initial conditions $θ (0) = Θ$ , $\dot{θ} (0) = 0$ give phase $ϕ = 0$ :

$θ (t) = Θ cos (ω t) = 0.100 cos (3.130 t) rad .$

(c) Find the maximum tangential speed $v_{m a x}$ of the bob.

Tangential speed $v = L \dot{θ}$ . Since $\dot{θ}_{m a x} = ω Θ$ ,

$v_{m a x} = L ω Θ = (1.00) (3.130) (0.100) = 0.313 m/s .$

(d) Find the maximum tangential acceleration $a_{m a x}$ of the bob.

Tangential acceleration $a = L \ddot{θ}$ . Since $\ddot{θ}_{m a x} = ω^{2} Θ$ ,

$a_{m a x} = L ω^{2} Θ = (1.00) (3.130)^{2} (0.100) = 0.979 m/ s^{2} .$

Simple Harmonic Motion (SHM) Basics\

Oscillation about equilibrium under restoring force $\propto$ displacement
Acceleration: $a = - ω^{2} x$ (points toward equilibrium)
Angular frequency: $ω = \frac{k}{m}$ (rad/s)

Mass-spring system\

Equation of motion: $\overset{x}{¨} + ω^{2} x = 0$
Solution: $x (t) = A cos (ω t + ϕ)$
Velocity and acceleration:
- $v (t) = - A ω sin (ω t + ϕ)$
- $a (t) = - ω^{2} x (t)$

Period and frequency in SHM\

Period: $T = \frac{2 π}{ω}$
Frequency: $f = \frac{1}{T}$
For mass-spring: $T = 2 π \frac{m}{k}$
For pendulum (small angle): $T = 2 π \frac{L}{g}$
$T$ and $f$ independent of amplitude $A$

Graphical representation of SHM\

Displacement: $x / A = cos (ω t + ϕ)$
Velocity: $v / (A ω) = - sin (ω t + ϕ)$
Phase relationships:
- Velocity leads displacement by $9 0^{\circ}$
- Acceleration opposite in phase to displacement

Simple pendulum (small-angle approximation)\

Restoring force: $F_{tan} \approx - m g \frac{s}{L}$
Equation of motion: $\overset{s}{¨} + \frac{g}{L} s = 0$
Angular frequency: $ω = \frac{g}{L}$
Solution: $s (t) = S cos (ω t + ϕ)$
Period: $T = 2 π \frac{L}{g}$

Example problem 1: Mass-spring system\

$ω = \frac{k}{m}$ , $T = \frac{2 π}{ω}$
$x (t) = A cos (ω t)$ for release from rest at $x = A$
Speed at $x$ : $v = \pm ω A^{2} - x^{2}$
Acceleration at $x$ : $a = - ω^{2} x$

Example problem 2: Simple pendulum\

$ω = \frac{g}{L}$ , $T = \frac{2 π}{ω}$
$θ (t) = Θ cos (ω t)$ for release from rest at $θ = Θ$
Maximum tangential speed: $v_{m a x} = L ω Θ$
Maximum tangential acceleration: $a_{m a x} = L ω^{2} Θ$

Key takeaways\

SHM: restoring force $\propto$ displacement, leads to sinusoidal motion
Acceleration always points toward equilibrium
Period/frequency depend on system parameters, not amplitude
Recognize SHM by force law, equations, or motion graphs
Understand phase relationships between $x$ , $v$ , and $a$

Next | 8.2 Energy and dynamics in SHM

Simple harmonic motion

A car’s suspension is a tuned bounce-and-return system that helps keep the wheels on the road.
Skyscrapers can sway in the wind; to reduce the motion, builders sometimes install large pendulums near the top - heavy swinging masses that counter the sway.
A phone’s vibration motor spins an off-center weight, which makes the phone’s casing oscillate in a quick, controlled way.

In SHM, the acceleration always points toward equilibrium and satisfies

$a = - ω^{2} x,$

where

$ω \equiv \frac{k}{m} (rad/s)$

is the angular frequency, $m$ is the mass, $k$ is the spring constant, and $x$ is the displacement from equilibrium.

Mass-spring derivation

Free‐body diagram:

Equation of motion. Apply Newton’s second law along the horizontal direction:

$\sum F_{x} = m a_{x}, - k x = m \overset{x}{¨} .$

Rearrange:

$\overset{x}{¨} + \frac{k}{m} x = 0.$

Define

$ω \equiv \frac{k}{m},$

so the differential equation becomes

$\overset{x}{¨} + ω^{2} x = 0.$

You don’t need to do the calculus to use SHM results in most problems. The derivatives here are included to show how the standard formulas fit together.

A standard solution to this equation is

$x (t) = A cos (ω t + ϕ),$

where $A$ is the amplitude (maximum displacement) and $ϕ$ is the phase constant (set by the initial conditions). Once $x (t)$ is known, velocity and acceleration come from differentiation:

$v (t) = \overset{x}{˙} (t) = - A ω sin (ω t + ϕ),$

$a (t) = \overset{x}{¨} (t) = - A ω^{2} cos (ω t + ϕ) = - ω^{2} x (t) .$

Thus:

Velocity leads displacement by a quarter-cycle (90°).
Acceleration is in phase with displacement but opposite in sign.

Period and frequency in SHM

$f = \frac{1}{T} .$

Since the displacement in SHM can be written

$x (t) = A cos (ω t + ϕ),$

the argument $ω t$ must increase by $2 π$ for one full cycle. That gives

$ω T = 2 π ⟹ T = \frac{2 π}{ω} .$

Because $ω = k / m$ for a mass-spring system (or $g / L$ for a small-angle pendulum), the period depends only on the system’s parameters:

$T_{spring} = 2 π \frac{m}{k}, T_{pendulum} = 2 π \frac{L}{g} .$

All SHM systems with the same $ω$ share the same $T$ and $f$ , regardless of amplitude $A$ .
Higher $ω$ (stronger restoring force or smaller inertia) $\Rightarrow$ shorter $T$ $\Rightarrow$ higher $f$ .
Frequency is measured in hertz (Hz), where $1 Hz = 1$ cycle/sec.

Graphical representation of SHM

Here the solid blue curve is $x / A = cos (ω t + ϕ)$ and the red dashed curve is $v / (A ω) = - sin (ω t + ϕ)$ .

Simple pendulum (small‐angle approximation)

Restoring force and equation of motion

At an angular displacement $θ$ , the component of the weight $m g$ acting along the arc is

$F_{tan} = - m g sin θ .$

For small angles $(θ ≪ 1)$ , $sin θ \approx θ$ . If $s$ is the arc length from equilibrium, then

$s = L θ ⟹ sin θ \approx θ = \frac{s}{L},$

so the tangential force becomes

$F_{tan} \approx - m g \frac{s}{L} .$

Apply Newton’s second law in the tangential direction:

$m \overset{s}{¨} = - m g \frac{s}{L} ⟹ \overset{s}{¨} + \frac{g}{L} s = 0.$

This matches the SHM form $\overset{s}{¨} + ω^{2} s = 0$ with

$ω = \frac{g}{L} (rad/s) .$

General solution

The displacement along the arc varies sinusoidally:

$s (t) = S cos (ω t + ϕ),$

where $S$ is the maximum arc displacement (amplitude) and $ϕ$ is the phase constant set by initial conditions. Differentiating,

$v (t) = \overset{s}{˙} = - S ω sin (ω t + ϕ), a (t) = \overset{s}{¨} = - ω^{2} S cos (ω t + ϕ) = - ω^{2} s (t) .$

Period and frequency

One full oscillation corresponds to the argument of the cosine increasing by $2 π$ :

$ω T = 2 π ⟹ T = \frac{2 π}{ω} = 2 π \frac{L}{g}, f = \frac{1}{T} .$

In the small-angle limit, the pendulum behaves like a mass-spring system, with $ω = g / L$ .

Example Problem 1

$ω = \frac{k}{m} = \frac{50.0}{0.200} = 250 = 15.81 rad/s, T = \frac{2 π}{ω} = \frac{2 π}{15.81} = 0.397 s .$

(b) Write the equation of motion $x (t)$ assuming $t = 0$ at release.

Released from rest at $x = A$ implies phase $ϕ = 0$ , so

$x (t) = A cos (ω t) = 0.100 cos (15.81 t) m .$

(c) Determine the block’s speed when it passes through $x = 0.060$ m.

In SHM,

$v^{2} = ω^{2} (A^{2} - x^{2}) .$

Thus

$v = \pm ω A^{2} - x^{2} = 15.81 0.0064 = 1.265 m/s .$

Since the block is moving toward equilibrium, take $v = + 1.27$ m/s.

(d) Determine the acceleration of the block at $x = 0.060$ m.

Acceleration in SHM is $a = - ω^{2} x$ :

$a = - (15.81)^{2} (0.060) = - 250 (0.060) = - 15.0 m/ s^{2} .$

The negative sign shows the acceleration is toward equilibrium (leftward).

Example Problem 2

For small angles,

$ω = \frac{g}{L} = \frac{9.80}{1.00} = 3.130 rad/s, T = \frac{2 π}{ω} = \frac{2 π}{3.130} = 2.01 s .$

(b) Write $θ (t)$ assuming $θ (0) = Θ = θ_{m a x}$ and released from rest.

Initial conditions $θ (0) = Θ$ , $\dot{θ} (0) = 0$ give phase $ϕ = 0$ :

$θ (t) = Θ cos (ω t) = 0.100 cos (3.130 t) rad .$

(c) Find the maximum tangential speed $v_{m a x}$ of the bob.

Tangential speed $v = L \dot{θ}$ . Since $\dot{θ}_{m a x} = ω Θ$ ,

$v_{m a x} = L ω Θ = (1.00) (3.130) (0.100) = 0.313 m/s .$

(d) Find the maximum tangential acceleration $a_{m a x}$ of the bob.

Tangential acceleration $a = L \ddot{θ}$ . Since $\ddot{θ}_{m a x} = ω^{2} Θ$ ,

$a_{m a x} = L ω^{2} Θ = (1.00) (3.130)^{2} (0.100) = 0.979 m/ s^{2} .$

Achievable AP Physics 1

8. Oscillations

Our AP Physics 1 course is currently in development and is a work-in-progress.