Oscillations in Mechanics: Energy, Damping, and Resonance

Energy in SHM

What “energy in SHM” means (and why it’s so useful)

In simple harmonic motion (SHM), the object’s position oscillates back and forth about an equilibrium point due to a restoring force proportional to displacement. You already know the “motion” side of SHM (sinusoids, angular frequency, etc.). The energy viewpoint is equally powerful because it lets you answer many questions without solving for x(t) explicitly.

The key idea is that in ideal SHM (no friction, no air resistance), the system continuously trades energy between:

• Kinetic energy (energy of motion)
• Potential energy (stored in a spring or in gravity)

The total mechanical energy stays constant, and that single fact can give you speeds, amplitudes, turning points, and more.

A common misconception is thinking energy is “constant at each point” in the motion. It’s the total energy that is constant, while kinetic and potential energies vary with time and position.

SHM reference relationships (needed for energy connections)

For a mass–spring oscillator in SHM, the standard kinematics are:

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

v(t) = -A\omega\sin(\omega t + \phi)

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

where:

• A is amplitude (maximum displacement)
• \omega is angular frequency
• \phi is phase constant

For a horizontal mass–spring system:

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

where k is the spring constant and m is the mass.

You do not always need these time functions for energy problems—but it helps to remember that speed is maximum at equilibrium and zero at turning points, matching the energy story.

Energy in a mass–spring SHM system

For a spring, the elastic potential energy (relative to equilibrium) is:

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

The kinetic energy of the mass is:

K = \frac{1}{2}mv^2

In ideal SHM, total mechanical energy is constant:

E = K + U_s

At maximum displacement x = \pm A, the speed is zero, so all energy is spring potential:

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

At equilibrium x = 0, spring potential is zero, so all energy is kinetic:

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

Setting these equal gives a famous and very testable result:

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

Using \omega^2 = k/m, you also get:

v_{\max} = \omega A

This relationship is an excellent “consistency check”: maximum speed grows linearly with amplitude and with angular frequency.

Energy as a function of position (speed without time)

Because E is constant, you can solve for speed at any displacement x without using trig. Start with:

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

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

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

Solve for v:

v^2 = \frac{k}{m}(A^2 - x^2)

and therefore:

v = \pm\omega\sqrt{A^2 - x^2}

Interpretation matters:

• The object is fastest at x = 0.
• The object stops at x = \pm A.
• The sign of v depends on which way it’s moving, but the energy approach naturally gives you speed magnitude.

A common student mistake is to plug in x = A and still expect a nonzero speed, forgetting that turning points have zero kinetic energy.

Energy-time behavior (what the graphs should look like)

Even if you never draw these on the exam, mentally picturing them helps:

• x(t) is cosine-like.
• v(t) is sine-like and shifted by a quarter period.
• U_s \propto x^2 so it oscillates at **twice** the frequency of x(t) and never goes negative.
• K \propto v^2 also oscillates at twice the frequency of x(t) and never goes negative.
• When U_s is maximum, K is minimum (zero), and vice versa.

This “twice the frequency” idea is commonly tested conceptually.

SHM energy for a simple pendulum (small-angle approximation)

For a simple pendulum of length L and bob mass m, the exact gravitational potential energy change depends on height. But for small angles, the motion is approximately SHM.

Let \theta be the angular displacement (in radians). For small angles:

\sin\theta \approx \theta

The tangential restoring force is approximately:

F_t \approx -mg\theta

Since arc length displacement is x = L\theta, the equation of motion becomes SHM with:

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

Energy-wise, for small angles the potential energy relative to the bottom can be approximated by a quadratic form:

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

If you prefer using linear displacement x = L\theta, substitute \theta = x/L:

U \approx \frac{1}{2}\frac{mg}{L}x^2

This looks exactly like spring potential energy with an “effective spring constant”:

k_{\text{eff}} = \frac{mg}{L}

That equivalence is why small-angle pendulum motion behaves like SHM and why the energy looks like a parabola in x.

Important limitation: this quadratic approximation fails at larger angles; then the motion is not exactly sinusoidal and the period is no longer independent of amplitude.

Worked example 1: speed at a given displacement (mass–spring)

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 oscillates with amplitude A = 0.10\,\text{m}. Find the speed when the block is at x = 0.060\,\text{m}.

Step 1: Use energy conservation.

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

Solve for v^2:

v^2 = \frac{k}{m}(A^2 - x^2)

Step 2: Substitute values.

v^2 = \frac{200}{0.50}\left(0.10^2 - 0.060^2\right)

Compute the bracket:

0.10^2 - 0.060^2 = 0.0100 - 0.0036 = 0.0064

Then:

v^2 = 400\times 0.0064 = 2.56

So:

v = 1.6\,\text{m/s}

(That’s the speed magnitude. The velocity sign depends on whether it’s moving toward or away from equilibrium.)

Worked example 2: connecting maximum speed and amplitude

Using the same system, compute \omega and confirm v_{\max} = \omega A.

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = \sqrt{400} = 20\,\text{rad/s}

Then:

v_{\max} = \omega A = 20\times 0.10 = 2.0\,\text{m/s}

You could also get this from energy at equilibrium:

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

The agreement is a good internal check.

Exam Focus

• Typical question patterns:
  • Given k, m, and A, find E, v_{\max}, or the speed at a specific x using energy conservation.
  • Interpret or sketch how K and U vary over time or with position (often conceptual, sometimes with a graph).
  • For pendulums, use the small-angle approximation to treat the energy as quadratic and connect to SHM.
• Common mistakes:
  • Mixing up amplitude A with instantaneous displacement x (they are not the same except at turning points).
  • Forgetting that turning points have v = 0 and equilibrium has maximum speed.
  • Using the small-angle pendulum formulas at large angles (where the SHM model breaks down).

Damped and Driven Oscillations

Why damping and driving matter

Real oscillators don’t oscillate forever. Friction, air resistance, internal material losses, and electrical resistance all remove mechanical energy and convert it into thermal energy. This is called damping.

At the same time, many systems are forced to keep oscillating by some external periodic input (a push every cycle, a motor, a periodic force). That’s a driven oscillator. The combination of driving and damping explains resonance—the huge response that can happen when you drive near the natural frequency.

In AP Physics C: Mechanics, you are expected to connect forces, differential equations, and energy ideas. Even when the calculus looks intimidating, the physics story is simple:

• Damping removes energy, shrinking amplitude.
• Driving adds energy, potentially balancing damping.
• Resonance is large amplitude when energy input is efficient.

Linear (viscous) damping model

A widely used model assumes the damping force is proportional to velocity and opposite the motion:

F_d = -bv

where b is a damping coefficient.

For a mass–spring oscillator with damping, Newton’s second law gives:

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

Define two useful parameters:

\omega_0 = \sqrt{\frac{k}{m}}

\beta = \frac{b}{2m}

Here \omega_0 is the natural angular frequency without damping, and \beta sets the decay rate.

Types of damping: underdamped, critically damped, overdamped

The behavior depends on how strong the damping is compared to the restoring effect.

Underdamped (oscillatory decay)

If damping is weak enough, the mass still oscillates, but the amplitude decays exponentially:

x(t) = A_0 e^{-\beta t}\cos(\omega_d t + \phi)

where the damped angular frequency is:

\omega_d = \sqrt{\omega_0^2 - \beta^2}

Key takeaways:

• The oscillation frequency is slightly lower than \omega_0.
• The envelope decays as e^{-\beta t}.

A misconception to avoid: students sometimes think damping changes frequency a lot. For light damping (common in many problems), \omega_d is close to \omega_0.

Critically damped (fastest return without oscillation)

At the boundary between oscillatory and non-oscillatory motion:

\beta = \omega_0

In terms of b, the critical damping coefficient for a mass–spring system is:

b_c = 2\sqrt{km}

Critically damped systems return to equilibrium as quickly as possible without overshooting.

Overdamped (slow, non-oscillatory return)

If:

\beta > \omega_0

the system does not oscillate; it creeps back to equilibrium more slowly than the critically damped case.

Real-world examples:

• Door closers are designed to be close to critically damped so the door doesn’t oscillate and doesn’t take forever to shut.
• Car suspension is designed to avoid sustained oscillations after hitting a bump (too little damping feels “bouncy”).

Energy decay in a damped oscillator

In an underdamped oscillator, the amplitude decays as:

A(t) = A_0 e^{-\beta t}

Because mechanical energy in SHM scales like amplitude squared, energy decays faster:

E(t) = E_0 e^{-2\beta t}

This is a high-yield conceptual point: amplitude decays with e^{-\beta t}, energy with e^{-2\beta t}.

Quality factor and “sharpness” of resonance

The quality factor Q is a dimensionless measure of how lightly damped an oscillator is. For light damping, a common definition is:

Q = \frac{\omega_0}{2\beta}

Higher Q means:

• slower energy loss per cycle
• narrower, sharper resonance peak in driven motion

You do not need to memorize every alternate definition if you understand the physical meaning: high Q systems ring for a long time.

Driven oscillations: adding an external periodic force

Now include a driving force, often modeled as:

F(t) = F_0\cos(\omega t)

The equation becomes:

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

After transient behavior dies out (the part depending on initial conditions), the system reaches a steady-state oscillation at the driving frequency \omega:

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

Here:

• A(\omega) is the frequency-dependent amplitude
• \delta is the phase lag (how much the response lags behind the driver)

This is crucial: in steady state, the oscillator moves at the driving frequency, not its natural frequency. The natural frequency shows up in how large the amplitude becomes.

Amplitude response curve (resonance)

For the damped, driven oscillator, the steady-state amplitude is:

A(\omega) = \frac{F_0}{\sqrt{(k - m\omega^2)^2 + (b\omega)^2}}

Interpretation:

• When \omega is very small, the mass follows the force quasi-statically and amplitude approaches F_0/k.
• When \omega is very large, inertia dominates and amplitude becomes small.
• Near the natural frequency, amplitude can become large: resonance.

With damping present, the amplitude stays finite even at resonance (because energy is continuously dissipated).

A common mistake is to assume maximum amplitude occurs exactly at \omega_0 for any damping. For light damping it is close, but the exact peak occurs slightly below \omega_0.

Phase difference between driving force and displacement

The phase lag \delta satisfies:

\tan\delta = \frac{b\omega}{k - m\omega^2}

Qualitative behavior:

• At low frequency, \delta is near 0: displacement roughly in phase with force.
• Near resonance, \delta is near \pi/2: displacement lags by about a quarter cycle.
• At high frequency, \delta approaches \pi: displacement nearly out of phase.

Students often confuse phase between force and velocity versus force and displacement. For power/energy transfer, velocity matters (because instantaneous power is P = Fv), but the standard phase formula above is about displacement relative to the driving force.

Power and energy flow in steady state

In steady state, the driver supplies energy each cycle, and damping removes energy each cycle. The average power dissipated by linear damping is:

P_{\text{avg}} = \frac{1}{2}b\omega^2 A^2

This result comes from the fact that the damping force is proportional to velocity and that the time average of v^2 over a cycle is proportional to \omega^2A^2.

Physical meaning:

• If you increase damping (larger b), more power is lost for the same amplitude.
• Near resonance, amplitude grows, so dissipated power can become large.

Notation and parameter reference (common in problems)

Different sources use slightly different symbols. Here’s a consistent map.

Worked example 1: amplitude decay and energy decay

A damped oscillator has \beta = 0.20\,\text{s}^{-1} and initial amplitude A_0 = 0.050\,\text{m}.

(a) Find the amplitude after t = 6.0\,\text{s}.

Use:

A(t) = A_0 e^{-\beta t}

Substitute:

A(6.0) = 0.050 e^{-0.20\times 6.0} = 0.050 e^{-1.2}

Numerically, e^{-1.2} \approx 0.301, so:

A(6.0) \approx 0.015\,\text{m}

(b) What fraction of the original mechanical energy remains?

Use:

\frac{E(t)}{E_0} = e^{-2\beta t}

So:

\frac{E(6.0)}{E_0} = e^{-2(0.20)(6.0)} = e^{-2.4}

With e^{-2.4} \approx 0.091, about 9\% of the energy remains.

The main conceptual point: energy drops much faster than amplitude because energy depends on amplitude squared.

Worked example 2: steady-state amplitude of a driven oscillator

A mass–spring system has m = 1.0\,\text{kg}, k = 100\,\text{N/m}, damping coefficient b = 6.0\,\text{kg/s}, and driving force amplitude F_0 = 10\,\text{N}. Find the steady-state amplitude when the driving frequency is \omega = 10\,\text{rad/s}.

Compute the denominator in:

A(\omega) = \frac{F_0}{\sqrt{(k - m\omega^2)^2 + (b\omega)^2}}

First:

k - m\omega^2 = 100 - 1.0\times 10^2 = 0

Then:

b\omega = 6.0\times 10 = 60

So:

A = \frac{10}{\sqrt{0^2 + 60^2}} = \frac{10}{60} = 0.167\,\text{m}

Notice what happened: the chosen \omega equals \omega_0 here because \omega_0 = \sqrt{100/1} = 10\,\text{rad/s}, so the spring and inertia terms cancel in the steady-state balance, leaving damping to set the amplitude.

Exam Focus

• Typical question patterns:
  • Identify whether motion is underdamped, critically damped, or overdamped from parameter relationships or qualitative graphs of x(t).
  • Use A(t) = A_0 e^{-\beta t} or E(t) = E_0 e^{-2\beta t} to compute decay after a given time.
  • For driven oscillations, compute amplitude A(\omega) and/or phase behavior near resonance; interpret resonance curves conceptually.
• Common mistakes:
  • Treating the driven steady-state motion as if it oscillates at \omega_0 instead of the driving frequency \omega.
  • Forgetting that damping limits amplitude at resonance; plugging \omega = \omega_0 and expecting “infinite amplitude.”
  • Confusing the exponential decay constant for amplitude with that for energy (mixing up e^{-\beta t} and e^{-2\beta t}).