AP Physics C Mechanics Unit 6 Notes: Learning Simple Harmonic Motion

Simple Harmonic Motion (Springs and Pendulums)

What “simple harmonic” means (concept first)

Simple harmonic motion (SHM) is a special kind of back-and-forth motion where the acceleration of the object is directly proportional to its displacement from equilibrium and points back toward equilibrium.

That sentence packs in the whole idea:

• Displacement from equilibrium means you measure position from the point where the net force would be zero (the balance point).
• Proportional to displacement means “twice as far” leads to “twice as much restoring acceleration.”
• Points back toward equilibrium means the acceleration always tries to undo the displacement.

Mathematically, the defining feature is:

a \propto -x

More explicitly, for SHM:

a = -\omega^2 x

Here x is displacement from equilibrium, a is acceleration, and \omega (omega) is the angular frequency (a constant for a given oscillator). This relationship is the reason SHM is so important: it turns a messy-looking physical situation into a motion you can describe exactly with sinusoidal functions.

Why SHM matters in mechanics

SHM shows up whenever you have a stable equilibrium with a restoring influence that is approximately linear for small displacements. In AP Physics C: Mechanics, SHM is a gateway topic because it ties together:

• Forces (Hooke’s law, gravity components)
• Energy (exchange between kinetic and potential)
• Differential equations (how forces create motion)
• Approximations (especially the small-angle approximation for pendulums)

A big idea: many real oscillations are only “approximately SHM,” but that approximation is often extremely useful.

The mass–spring oscillator (horizontal)

A spring obeys Hooke’s law when it is not stretched or compressed too far:

F_s = -kx

• k is the spring constant (stiffness).
• x is displacement from equilibrium length (for a horizontal spring, equilibrium is usually the unstretched length if no other forces act horizontally).
• The negative sign indicates the force points opposite the displacement.

Now apply Newton’s 2nd law to a mass m attached to the spring on a frictionless surface:

\sum F = ma

The only horizontal force is the spring force:

-kx = ma

So,

a = -\frac{k}{m}x

Compare that to the SHM definition a = -\omega^2 x, and you identify:

\omega = \sqrt{\frac{k}{m}}

So a stiffer spring (larger k) makes the oscillation “faster,” while a larger mass makes it “slower.”

Vertical spring: where students often get confused

With a vertical spring, gravity shifts the equilibrium position. If the mass hangs at rest, then at equilibrium:

k x_{eq} = mg

The oscillation happens around that shifted equilibrium. The key learning point is:

• If you define x as displacement from the equilibrium position (not from the unstretched spring), the motion is still SHM with the same \omega = \sqrt{k/m}.

A common mistake is trying to put mg into the SHM equation as if it changes the frequency. Gravity changes the equilibrium stretch, not the oscillation frequency (assuming the spring stays linear).

The simple pendulum (small angles)

A simple pendulum is a point mass (bob) on a massless string of length L. The restoring influence comes from gravity.

If the pendulum is displaced by an angle \theta from vertical, the tangential component of gravity is:

F_t = -mg\sin\theta

The minus sign means it points back toward equilibrium (toward \theta = 0).

The motion along the arc is rotational. The torque about the pivot is:

\tau = -mgL\sin\theta

For a point mass, the moment of inertia is:

I = mL^2

And rotational dynamics gives:

\sum \tau = I\alpha

So,

-mgL\sin\theta = mL^2 \alpha

Divide by mL^2:

\alpha = -\frac{g}{L}\sin\theta

Since \alpha = d^2\theta/dt^2, the exact equation is not SHM because of the \sin\theta.

Small-angle approximation (why SHM works for pendulums)

For sufficiently small angles (in radians), you can use:

\sin\theta \approx \theta

Then the equation becomes:

\alpha = -\frac{g}{L}\theta

That is the SHM form \theta'' = -\omega^2\theta with:

\omega = \sqrt{\frac{g}{L}}

So the pendulum behaves like SHM only when the amplitude is small enough that \sin\theta \approx \theta is valid.

Important conceptual takeaway: The “restoring force proportional to displacement” requirement is satisfied only approximately for a pendulum.

Energy view of SHM (applies to both springs and pendulums)

In SHM without nonconservative forces (no friction/air drag), mechanical energy is conserved and sloshes between kinetic energy and potential energy.

For a mass–spring oscillator, the spring potential energy is:

U_s = \frac{1}{2}kx^2

Total energy is constant:

E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2

At maximum displacement (amplitude), speed is zero; at equilibrium, speed is maximum.

For a pendulum, the gravitational potential energy relative to the lowest point is:

U_g = mgL(1-\cos\theta)

For small angles, you can approximate \cos\theta \approx 1-\theta^2/2, giving:

U_g \approx \frac{1}{2}mgL\theta^2

That quadratic form is one reason small-angle pendulums behave like SHM: the potential energy becomes approximately quadratic, just like a spring.

Worked example 1 (spring): maximum speed from energy

A block of mass m = 0.50\ \text{kg} on a frictionless surface is attached to a spring with k = 200\ \text{N/m}. It is pulled to amplitude A = 0.10\ \text{m} and released from rest.

Goal: Find the maximum speed.

At amplitude, all energy is spring potential:

E = \frac{1}{2}kA^2

At equilibrium, all energy is kinetic:

E = \frac{1}{2}mv_{max}^2

Set them equal:

\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2

Solve:

v_{max} = A\sqrt{\frac{k}{m}}

Plug in:

v_{max} = 0.10\sqrt{\frac{200}{0.50}}

v_{max} = 0.10\sqrt{400}

v_{max} = 0.10(20) = 2.0\ \text{m/s}

Notice how the factor \sqrt{k/m} is exactly \omega, so another common form is v_{max} = A\omega.

Worked example 2 (pendulum): when SHM is a valid model

A pendulum has length L = 1.0\ \text{m} and is released from rest at \theta_0 = 0.10\ \text{rad}.

Because 0.10\ \text{rad} is small, the approximation \sin\theta \approx \theta is reasonable, so treating it as SHM is appropriate in AP-style problems.

Its angular frequency is:

\omega = \sqrt{\frac{g}{L}}

With g = 9.8\ \text{m/s}^2 and L = 1.0\ \text{m}:

\omega = \sqrt{9.8} \approx 3.13\ \text{rad/s}

If the angle were much larger (for example, near 1\ \text{rad}), the period would deviate from the SHM prediction, and you should not blindly use the small-angle formulas.

Exam Focus

• Typical question patterns
  • Given a spring–mass system or pendulum, identify whether it performs SHM and write the restoring-force/torque relationship.
  • Use energy conservation to find speed at a position, or to relate amplitude to total energy.
  • For vertical springs, find equilibrium extension first, then analyze oscillations about equilibrium.
• Common mistakes
  • Using x measured from the unstretched length in a vertical spring SHM equation instead of measuring from equilibrium.
  • Treating a pendulum as SHM at large angles (forgetting that \sin\theta \approx \theta is an approximation).
  • Confusing “restoring force exists” with “restoring force proportional to displacement” (only the latter guarantees SHM).

Period and Frequency

What period and frequency represent

When something oscillates, you need language to describe how fast it repeats.

• The period T is the time for one full cycle (units: seconds).
• The frequency f is the number of cycles per second (units: hertz, where 1\ \text{Hz} = 1\ \text{s}^{-1}).

They are reciprocals:

f = \frac{1}{T}

In calculus-based physics, a third quantity is especially useful:

• The angular frequency \omega (units: rad/s), which connects oscillations to circular motion and makes the differential equation solution clean.

These are related by:

\omega = 2\pi f

and therefore:

\omega = \frac{2\pi}{T}

Why AP Physics emphasizes \omega

In SHM, acceleration is tied directly to displacement through \omega:

a = -\omega^2 x

So if you know \omega, you immediately know how “strong” the curvature of the motion is. Also, solutions like x(t) = A\cos(\omega t + \phi) naturally use \omega.

Period for a mass–spring system

From the spring SHM identification \omega = \sqrt{k/m}, the period is:

T = \frac{2\pi}{\omega}

So,

T = 2\pi\sqrt{\frac{m}{k}}

Key implications:

• Increasing m increases T (slower oscillations).
• Increasing k decreases T (faster oscillations).
• For ideal SHM, period does not depend on amplitude A.

That last point is often tested conceptually: if the amplitude doubles for an ideal spring, the period stays the same.

Period for a simple pendulum (small-angle)

From \omega = \sqrt{g/L} (valid for small angles), the period is:

T = 2\pi\sqrt{\frac{L}{g}}

Key implications:

• Longer L means larger T (slower swing).
• Larger g means smaller T (faster swing).
• For the small-angle model, T does not depend on mass or amplitude.

Students often find the “no mass dependence” surprising, but it follows from the way inertia and gravity both scale with m and cancel.

Notation reference (common forms you must recognize)

Worked example 1: compare two springs

Two different setups:

• System 1: m_1 = 1.0\ \text{kg}, k_1 = 100\ \text{N/m}
• System 2: m_2 = 1.0\ \text{kg}, k_2 = 400\ \text{N/m}

Find the ratio T_2/T_1.

Use T = 2\pi\sqrt{m/k}. The constants and equal masses cancel in the ratio:

\frac{T_2}{T_1} = \sqrt{\frac{k_1}{k_2}} = \sqrt{\frac{100}{400}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

So the stiffer spring oscillates with half the period.

Worked example 2: pendulum on a different planet (conceptual + calculation)

A pendulum clock calibrated on Earth (with g = 9.8\ \text{m/s}^2) is taken to a location where g = 3.7\ \text{m/s}^2, but its length L is unchanged.

From T = 2\pi\sqrt{L/g}, the period scales like 1/\sqrt{g}. So:

\frac{T_{new}}{T_{Earth}} = \sqrt{\frac{g_{Earth}}{g_{new}}} = \sqrt{\frac{9.8}{3.7}}

\frac{T_{new}}{T_{Earth}} \approx \sqrt{2.65} \approx 1.63

The pendulum swings more slowly, so the clock runs slow. This is a classic “period depends on g” application.

Exam Focus

• Typical question patterns
  • Compute T or f for a spring or pendulum and interpret how changing parameters affects the oscillation.
  • Given a graph of x(t) or a(t), extract T, f, and \omega.
  • Use proportional reasoning (ratios) to avoid heavy algebra when a parameter changes.
• Common mistakes
  • Mixing up \omega and f (forgetting the factor of 2\pi).
  • Using degrees instead of radians when connecting pendulum angle to SHM equations.
  • Assuming “bigger amplitude means bigger period” for ideal SHM (true for real pendulums at large angles, but not in the small-angle SHM model).

Differential Equation of SHM

Why a differential equation appears at all

In mechanics, motion is determined by Newton’s 2nd law:

\sum F = m\frac{d^2x}{dt^2}

Because acceleration is the second time derivative of position, any force law gives you a differential equation for x(t).

SHM is special because the restoring force is linear:

F = -kx

A linear force produces a linear differential equation with sinusoidal solutions you can write down exactly.

Deriving the SHM differential equation (spring)

Start with Newton’s 2nd law for a mass–spring system (with x measured from equilibrium):

m\frac{d^2x}{dt^2} = -kx

Rearrange:

\frac{d^2x}{dt^2} + \frac{k}{m}x = 0

Define:

\omega^2 = \frac{k}{m}

Then the standard SHM equation is:

\frac{d^2x}{dt^2} + \omega^2 x = 0

This equation is the core mathematical statement of SHM.

Deriving the SHM differential equation (pendulum, small-angle)

From rotational dynamics, you reached:

\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta

For small angles, use \sin\theta \approx \theta:

\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0

So for a simple pendulum in the small-angle regime:

\omega^2 = \frac{g}{L}

Again, it matches the SHM template.

Solving the SHM differential equation (what the solution means)

The differential equation

\frac{d^2x}{dt^2} + \omega^2 x = 0

is solved by sinusoidal functions. A standard position function is:

x(t) = A\cos(\omega t + \phi)

• A is the amplitude (maximum displacement).
• \phi is the phase constant, determined by initial conditions.
• The cosine form is conventional, but sine is equally valid.

To see how it works, compute derivatives:

v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)

a(t) = \frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t + \phi)

Since x(t) = A\cos(\omega t + \phi), you immediately get:

a(t) = -\omega^2 x(t)

That is exactly the SHM defining relationship.

Interpreting phase and initial conditions (how to build the correct function)

The hardest part for many students is not the calculus, but choosing A and \phi correctly from a physical description.

Typical initial condition types:

1. Released from rest at maximum displacement
   • If at t = 0 you have x(0) = A and v(0) = 0, then a clean choice is \phi = 0 in the cosine form:

x(t) = A\cos(\omega t)

2. Passes through equilibrium moving positive
   • If at t = 0 you have x(0) = 0 and v(0) > 0, then cosine with \phi = -\pi/2 works, but it is often simpler to switch to sine:

x(t) = A\sin(\omega t)

The physics is the same; you are just choosing the phase origin.

Amplitude sets energy, not period

A very testable conceptual point: in ideal SHM, \omega is set by system parameters (like k and m, or g and L), while A is set by how you start the motion.

For a spring,

E = \frac{1}{2}kA^2

So doubling A quadruples energy, but does not change T.

Connection between acceleration, velocity, and displacement

Once you know the sinusoidal solution, several useful “at a glance” facts follow:

• At maximum displacement, speed is zero and acceleration magnitude is maximum.
• At equilibrium, acceleration is zero and speed magnitude is maximum.

From the expressions:

v_{max} = A\omega

a_{max} = A\omega^2

These are frequent targets in AP problems because they combine conceptual understanding with clean algebra.

Worked example 1: write x(t) from physical initial conditions

A mass–spring system has m = 0.20\ \text{kg}, k = 50\ \text{N/m}. It is pulled to the right 0.030\ \text{m} and released from rest at t = 0.

Step 1: Compute \omega.

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.20}} = \sqrt{250}

\omega \approx 15.8\ \text{rad/s}

Step 2: Translate the initial condition into A and \phi.

“Pulled to the right 0.030 m and released from rest” means:

• x(0) = +0.030\ \text{m}
• v(0) = 0

That is exactly “released from rest at maximum displacement,” so choose cosine with \phi = 0 and A = 0.030\ \text{m}:

x(t) = 0.030\cos(15.8t)

A common mistake is putting A into the argument or treating \omega like it is in Hz; the argument must be in radians, so you need \omega t with \omega in rad/s.

Worked example 2: check that a proposed solution satisfies the differential equation

Suppose someone claims an SHM solution is:

x(t) = A\cos(\omega t + \phi)

Differentiate twice:

\frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t + \phi)

But A\cos(\omega t + \phi) = x(t), so:

\frac{d^2x}{dt^2} = -\omega^2 x

Rearrange:

\frac{d^2x}{dt^2} + \omega^2 x = 0

So the proposed solution satisfies the SHM differential equation.

This kind of “verify the solution” reasoning is helpful if an exam question gives you a function and asks what differential equation it solves, or what \omega is.

Exam Focus

• Typical question patterns
  • Start from \sum F = ma or \sum \tau = I\alpha and derive the SHM differential equation, then identify \omega.
  • Given initial conditions (released from rest, passing equilibrium with a given speed), write x(t) or \theta(t).
  • Use derivatives of x(t) to find v(t), a(t), and maximum values like v_{max}.
• Common mistakes
  • Forgetting that the SHM form requires displacement measured from equilibrium (especially for vertical springs).
  • Using the pendulum equation \theta'' + (g/L)\sin\theta = 0 as if it were SHM without explicitly invoking the small-angle approximation.
  • Confusing phase constant \phi with frequency information; \phi shifts the graph in time but does not change T or \omega.