Damped and Driven Oscillations

🔄 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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📉 Damped Oscillations

Part 4 of 7 — Friction and Decay

Damping Force

$F_{\text{damp}} = -b\frac{dx}{dt}$

The equation of motion becomes:

$m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0$

Solution: Underdamped Case

$x(t) = Ae^{-\gamma t}\cos(\omega' t + \phi)$

where $\gamma = \dfrac{b}{2m}$ and $\omega' = \sqrt{\omega_0^2 - \gamma^2}$.

Three Damping Regimes

🔑 Critical damping is used in car suspensions — the fastest return to equilibrium without overshooting.

📝 Worked Example — Amplitude Decay of an Underdamped Oscillator

A damped oscillator has $m = 0.50\,\text{kg}$, $k = 200\,\text{N/m}$, and damping constant $b = 0.50\,\text{kg/s}$. Find the decay constant $\gamma$, the damped frequency $\omega'$, and how long until the amplitude falls to half its initial value.

Step 1 — Natural frequency and decay constant.

$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = 20\,\text{rad/s}, \qquad \gamma = \frac{b}{2m} = \frac{0.50}{2(0.50)} = 0.50\,\text{s}^{-1}$

Step 2 — Damped angular frequency. Since $\gamma \ll \omega_0$, the system is underdamped:

$\omega' = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{20^2 - 0.5^2} = \sqrt{400 - 0.25} \approx 19.99\,\text{rad/s}$

Step 3 — Amplitude envelope. The amplitude is $A(t) = A_0 e^{-\gamma t}$. Set it to $A_0/2$ and solve:

$e^{-\gamma t} = \tfrac{1}{2} \;\Rightarrow\; -\gamma t = \ln\tfrac{1}{2} \;\Rightarrow\; t = \frac{\ln 2}{\gamma} = \frac{0.693}{0.50} \approx 1.39\,\text{s}$

🔑 The cosine keeps oscillating at nearly $\omega_0$, but the $e^{-\gamma t}$ envelope shrinks the amplitude. Light damping barely shifts the frequency.

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🔊 Driven Oscillations & Resonance

Part 5 of 7 — Forced Vibrations

Driven Oscillation

Apply a periodic driving force $F(t) = F_0\cos(\omega_d t)$:

$m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = F_0\cos(\omega_d t)$

Steady-state solution: $x(t) = A(\omega_d)\cos(\omega_d t - \delta)$

Resonance

$A(\omega_d) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega_d^2)^2 + (b\omega_d/m)^2}}$

The amplitude is maximum near $\omega_d = \omega_0$ (the natural frequency).

🔑 Resonance = the driving frequency matches the natural frequency, producing maximum energy transfer and the largest amplitude.

📝 Worked Example — Amplitude On and Off Resonance

A driven oscillator has $m = 1.0\,\text{kg}$, $k = 100\,\text{N/m}$, $b = 2.0\,\text{kg/s}$, and a driving force amplitude $F_0 = 5.0\,\text{N}$. Compare the steady-state amplitude at resonance ($\omega_d = \omega_0$) with a low-frequency drive ($\omega_d \to 0$).

Step 1 — Natural frequency.

$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1.0}} = 10\,\text{rad/s}$

Step 2 — Amplitude at resonance. At $\omega_d = \omega_0$ the $(\omega_0^2 - \omega_d^2)$ term vanishes, leaving:

$A_{\text{res}} = \frac{F_0/m}{b\omega_0/m} = \frac{F_0}{b\omega_0} = \frac{5.0}{(2.0)(10)} = 0.25\,\text{m}$

Step 3 — Amplitude at very low frequency. As $\omega_d \to 0$, the denominator approaches $\omega_0^2$:

$A_{0} = \frac{F_0/m}{\omega_0^2} = \frac{F_0}{k} = \frac{5.0}{100} = 0.050\,\text{m}$

Result: Driving at resonance gives an amplitude $0.25/0.050 = 5\times$ larger than the slow ("static") response — and smaller damping $b$ would make the peak even sharper.

🔑 The resonance peak height is set by the damping: smaller $b$ → taller, narrower peak (higher quality factor $Q = \omega_0 m / b$).

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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 an undamped 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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