Simple Harmonic Motion

🔄 Simple Harmonic Motion

Part 1 of 7 — Springs and Pendulums

Defining SHM

Simple harmonic motion occurs when the restoring force is proportional to displacement and directed toward equilibrium:

$F = -kx$

This leads to the differential equation:

$m\frac{d^2x}{dt^2} = -kx \quad\Longrightarrow\quad \frac{d^2x}{dt^2} = -\omega^2 x$

Solution: $x(t) = A\cos(\omega t + \phi)$

Key Quantities

Energy in SHM

$E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2$

🔑 Total mechanical energy is constant in SHM (no friction).

📝 Worked Example — Verifying the SHM Differential Equation

Show that $x(t) = A\cos(\omega t + \phi)$ solves $\dfrac{d^2x}{dt^2} = -\omega^2 x$, and use it to find the speed at $x = A/2$.

Step 1 — Differentiate once for velocity.

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

Step 2 — Differentiate again for acceleration.

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

Step 3 — Compare with the original. Since $x = A\cos(\omega t + \phi)$, we can substitute:

$a = -\omega^2\big[A\cos(\omega t + \phi)\big] = -\omega^2 x \quad✅$

The proposed function satisfies the equation for any amplitude $A$ and phase $\phi$.

Step 4 — Speed at $x = A/2$ using energy. Conservation of energy gives $\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2$, so:

$v = \omega\sqrt{A^2 - x^2} = \omega\sqrt{A^2 - (A/2)^2} = \omega A\sqrt{1 - \tfrac{1}{4}} = \frac{\sqrt{3}}{2}\,\omega A \approx 0.87\,v_{\max}$

🔑 Differentiating the position twice always returns $-\omega^2$ times the original — that is the signature of SHM.

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📐 SHM Kinematics

Part 2 of 7 — Position, Velocity, Acceleration

The Three Equations

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

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

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

Maximum Values

• Position: $\lvert x \rvert_{\max} = A$
• Velocity: $\lvert v \rvert_{\max} = A\omega$
• Acceleration: $\lvert a \rvert_{\max} = A\omega^2$

🔑 Velocity leads position by $\pi/2$. Acceleration leads velocity by $\pi/2$. Acceleration is $\pi$ out of phase with position (they point opposite ways).

📝 Worked Example — From Position Function to Velocity and Acceleration

A particle moves as $x(t) = 0.10\cos(8t)\,\text{m}$ (SI units, so $\omega = 8\,\text{rad/s}$ and $\phi = 0$). Find the velocity and acceleration at $t = \dfrac{\pi}{16}\,\text{s}$.

Step 1 — Differentiate for velocity.

$v(t) = \frac{dx}{dt} = -0.10(8)\sin(8t) = -0.80\sin(8t)\,\text{m/s}$

Step 2 — Differentiate again for acceleration.

$a(t) = \frac{dv}{dt} = -0.80(8)\cos(8t) = -6.4\cos(8t)\,\text{m/s}^2$

Step 3 — Evaluate at $t = \pi/16$. Note $8t = 8\cdot\dfrac{\pi}{16} = \dfrac{\pi}{2}$, so $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$:

$v = -0.80(1) = -0.80\,\text{m/s}, \qquad a = -6.4(0) = 0\,\text{m/s}^2$

Interpretation: At this instant the particle is at $x = 0.10\cos(\pi/2) = 0$ — the equilibrium point — so it has maximum speed ($\lvert v \rvert = A\omega = 0.80\,\text{m/s}$) and zero acceleration, exactly as predicted.

🔑 Each derivative shifts the phase by $\pi/2$: cosine → $-$sine → $-$cosine.

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🕐 Pendulums

Part 3 of 7 — Simple and Physical Pendulums

Simple Pendulum (small angle)

For small angles ($\theta < 15^\circ$):

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

Note: the period is independent of mass and independent of amplitude (for small angles).

Physical Pendulum

Any rigid body oscillating about a pivot:

$T = 2\pi\sqrt{\frac{I}{mgd}}$

where $I$ is the moment of inertia about the pivot and $d$ is the distance from the pivot to the center of mass.

Torsional Oscillator

$T = 2\pi\sqrt{\frac{I}{\kappa}}$

where $\kappa$ is the torsional constant (restoring torque $= -\kappa\theta$).

📝 Worked Example — Deriving the Pendulum Period from Torque

Start from the rotational form of Newton's second law, $\tau = I\alpha$, to show a simple pendulum undergoes SHM and find its period.

Step 1 — Write the restoring torque. For a bob of mass $m$ at length $L$, gravity supplies a torque about the pivot:

$\tau = -mgL\sin\theta$

Step 2 — Apply $\tau = I\alpha$ with $I = mL^2$ and $\alpha = \dfrac{d^2\theta}{dt^2}$.

$mL^2\frac{d^2\theta}{dt^2} = -mgL\sin\theta$

Step 3 — Use the small-angle approximation $\sin\theta \approx \theta$.

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

This has the SHM form $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$ with $\omega = \sqrt{g/L}$.

Step 4 — Convert to a period.

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

Numeric check: a $1.0\,\text{m}$ pendulum on Earth has $T = 2\pi\sqrt{1.0/9.8} \approx 2.0\,\text{s}$.

🔑 SHM emerges only because of the small-angle approximation $\sin\theta \approx \theta$. For large swings the motion is periodic but no longer simple harmonic.

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🔍 Identifying SHM

Part 4 of 7 — Restoring Forces, Spring Combinations, and Vertical Springs

What Makes Motion "Simple Harmonic"?

SHM is a special case of periodic motion. Two ideas define it:

• An equilibrium position is where the net force on the object is zero.
• A restoring force points opposite the displacement from that equilibrium.

The motion is simple harmonic when the restoring force is proportional to the displacement from equilibrium:

$F_{\text{net}} = -k_{\text{eff}}\,(x - x_{\text{eq}})$

Newton's second law then gives the SHM differential equation, with the displacement $x - x_{\text{eq}}$ playing the role of $x$:

$\frac{d^2x}{dt^2} = -\frac{k_{\text{eff}}}{m}(x - x_{\text{eq}}) \quad\Longrightarrow\quad \omega = \sqrt{\frac{k_{\text{eff}}}{m}}$

The Three-Step Test

1. Find equilibrium: set $F_{\text{net}}(x) = 0$ and solve for $x_{\text{eq}}$.
2. Check proportionality: is $F_{\text{net}}$ a constant times $-(x - x_{\text{eq}})$? If the force grows like $x^3$, or stays constant in size, the motion is periodic at best but not SHM.
3. Read off $k_{\text{eff}}$ (the magnitude of the slope of $F$ versus $x$) and compute $\omega = \sqrt{k_{\text{eff}}/m}$.

The same test works for rotation: a restoring torque $\tau = -\kappa\theta$ gives $I\,\dfrac{d^2\theta}{dt^2} = -\kappa\theta$, so $\omega = \sqrt{\kappa/I}$. That is the torsion pendulum from Part 3.

🔑 A constant force (such as gravity on a hanging mass) only shifts the equilibrium position. It does not change $k_{\text{eff}}$, so it does not change $\omega$.

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📝 Worked Example — Two Springs and a Vertical Hang

(a) A block between two springs. A $0.50\,\text{kg}$ block on a frictionless table is attached to a wall on the left by a spring with $k_1 = 120\,\text{N/m}$ and to a wall on the right by a spring with $k_2 = 80\,\text{N/m}$. Find the period.

Displace the block by $x$ to the right. The left spring stretches by $x$ and pulls left with $k_1x$; the right spring compresses by $x$ and pushes left with $k_2x$. Both forces point back toward equilibrium, and both springs share the same displacement, so they act in parallel:

$F_{\text{net}} = -(k_1 + k_2)x \;\Rightarrow\; k_{\text{eq}} = 200\,\text{N/m}$

$\omega = \sqrt{\frac{200}{0.50}} = 20\,\text{rad/s}, \qquad T = \frac{2\pi}{20} \approx 0.314\,\text{s}$

(b) The same springs joined end to end (series). Now the two springs carry the same force and their stretches add:

$\frac{1}{k_{\text{eq}}} = \frac{1}{120} + \frac{1}{80} \;\Rightarrow\; k_{\text{eq}} = \frac{(120)(80)}{200} = 48\,\text{N/m}$

$\omega = \sqrt{\frac{48}{0.50}} = \sqrt{96} \approx 9.8\,\text{rad/s}, \qquad T \approx 0.64\,\text{s}$

The series combination is softer than either spring alone, so the period is longer.

(c) A vertical spring. A $0.20\,\text{kg}$ block hangs from a spring with $k = 50\,\text{N/m}$ ($g = 10\,\text{m/s}^2$). At equilibrium the spring force balances gravity:

$k\,\Delta y = mg \;\Rightarrow\; \Delta y = \frac{(0.20)(10)}{50} = 0.040\,\text{m}$

Measure $y$ from this hanging position. A further displacement $y$ changes the spring force by $-ky$ while gravity stays $mg$, so $F_{\text{net}} = -ky$. The block oscillates about the hanging position with $\omega = \sqrt{50/0.20} \approx 15.8\,\text{rad/s}$, exactly as it would horizontally.

🔑 Parallel springs add constants ($k_{\text{eq}} = k_1 + k_2$); series springs add inverses and give $k_{\text{eq}}$ smaller than the smallest spring. Gravity moves the center of a vertical oscillation but leaves the period alone.

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🔊 Natural Frequency, Resonance, and SHM Graphs

Part 5 of 7 — Natural Frequency, Resonance, and SHM Graphs

Natural Frequency

The natural frequency of a system is the frequency at which it oscillates on its own after being displaced from equilibrium and released:

Convert with $f_0 = \dfrac{\omega_0}{2\pi}$ and $T = \dfrac{1}{f_0}$.

Resonance

Now push the system with a sinusoidal external force. If the push repeats at the natural frequency, each push arrives in step with the motion and adds energy every cycle, so the amplitude grows. This is resonance:

• Resonance occurs when the external force is exerted at the natural frequency of the system.
• Resonance increases the amplitude of the oscillation.

Everyday examples: pushing a child on a swing once per swing, a singer's note shattering a glass, a washing machine shaking hardest at one spin speed.

The AP Physics C exam treats resonance qualitatively; formulas for the driven amplitude are not tested.

Amplitude Does Not Change the Period

For SHM, $T = 2\pi\sqrt{m/k}$ (or $2\pi\sqrt{L/g}$) contains no amplitude. Pulling a spring farther before release raises the energy $\tfrac{1}{2}kA^2$ and the maximum speed $A\omega$, but the period stays the same.

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📝 Worked Example — Reading an SHM Graph

A graph of position versus time for a block on a spring shows: $x = 0$ at $t = 0$ with a positive slope, a first maximum of $x = 0.05\,\text{m}$ at $t = 0.25\,\text{s}$, and a return to $x = 0$ at $t = 0.50\,\text{s}$. Write $x(t)$ and find $v_{\max}$.

Step 1 — Amplitude. The largest displacement is $A = 0.05\,\text{m}$.

Step 2 — Period. Going from equilibrium to the maximum is one quarter of a cycle, so $T = 4(0.25) = 1.0\,\text{s}$ and

$\omega = \frac{2\pi}{T} = 2\pi\,\text{rad/s}$

Step 3 — Phase. The block starts at equilibrium moving in the $+x$ direction. A sine starts that way:

$x(t) = 0.05\sin(2\pi t) = 0.05\cos\!\left(2\pi t - \tfrac{\pi}{2}\right)\,\text{m}$

so in the form $A\cos(\omega t + \phi)$, the phase constant is $\phi = -\pi/2$.

Step 4 — Maximum speed. The slope of the graph is the velocity, and it is steepest at the equilibrium crossings:

$v_{\max} = A\omega = 0.05(2\pi) \approx 0.31\,\text{m/s}$

Reading other features from the graph:

🔑 Read $A$ from the peak height, $T$ from the time between matching points, and $\phi$ from where the graph starts and which way it is heading.

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🛠️ Oscillations Workshop

Part 6 of 7 — AP Physics C Strategies

Oscillation Problem Types

Worked Example

A $0.5\,\text{kg}$ mass on a spring ($k = 200\,\text{N/m}$) is pulled $0.1\,\text{m}$ and released.

$\omega = \sqrt{200/0.5} = 20\,\text{rad/s}$

$T = 2\pi/20 = 0.314\,\text{s}$

$v_{\max} = 0.1 \times 20 = 2\,\text{m/s}$

$E = \tfrac{1}{2}(200)(0.1)^2 = 1\,\text{J}$

📝 Worked Example — Maximum Speed by Differentiation

For the same oscillator, $x(t) = 0.1\cos(20t)\,\text{m}$. Confirm the maximum speed using calculus and find the first time the mass reaches it.

Step 1 — Velocity from the derivative.

$v(t) = \frac{dx}{dt} = -0.1(20)\sin(20t) = -2\sin(20t)\,\text{m/s}$

Step 2 — Locate the maximum speed. Speed is greatest where $\lvert\sin(20t)\rvert = 1$. The acceleration is the next derivative:

$a(t) = \frac{dv}{dt} = -40\cos(20t)$

Setting $a = 0$ gives $\cos(20t) = 0$, i.e. $20t = \dfrac{\pi}{2}$, so the first time is $t = \dfrac{\pi}{40} \approx 0.0785\,\text{s}$.

Step 3 — Evaluate. At that instant $\sin(20t) = \sin(\pi/2) = 1$, so $\lvert v \rvert = 2\,\text{m/s}$, matching $v_{\max} = A\omega = 0.1(20) = 2\,\text{m/s}$. ✅

Energy cross-check: $\tfrac{1}{2}mv_{\max}^2 = \tfrac{1}{2}(0.5)(2)^2 = 1\,\text{J} = E$ — all energy is kinetic at $x = 0$.

🔑 Maximizing speed means setting $a = dv/dt = 0$, which happens exactly at the equilibrium crossing.

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📋 Oscillations Review

Part 7 of 7 — Comprehensive Summary

Master Formula Sheet

🔑 SHM = any system with $F \propto -x$. Recognize the pattern, then apply the formulas.

📝 Worked Example — Building Energy from the Equation of Motion

Starting from $m\dfrac{d^2x}{dt^2} = -kx$, derive the conserved energy of a spring oscillator.

Step 1 — Multiply both sides by $v = \dfrac{dx}{dt}$.

$m\frac{dv}{dt}\,v = -kx\frac{dx}{dt}$

Step 2 — Recognize each side as a derivative. Using $\dfrac{d}{dt}\big(\tfrac{1}{2}v^2\big) = v\dfrac{dv}{dt}$ and $\dfrac{d}{dt}\big(\tfrac{1}{2}x^2\big) = x\dfrac{dx}{dt}$:

$\frac{d}{dt}\left(\tfrac{1}{2}mv^2\right) = -\frac{d}{dt}\left(\tfrac{1}{2}kx^2\right)$

Step 3 — Combine and integrate.

$\frac{d}{dt}\left(\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2\right) = 0 \;\Rightarrow\; \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = E = \text{constant}$

Step 4 — Identify the constant. At the turning point $x = A$, $v = 0$, so $E = \tfrac{1}{2}kA^2$. This also yields $v = \omega\sqrt{A^2 - x^2}$.

🔑 The "multiply by velocity and integrate" trick converts an equation of motion into an energy-conservation statement — a recurring AP Physics C calculus technique.

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