David Bauer Physics & Astronomy · UCLA

Physics 5B Chapter 17

Damped and Driven Oscillations

Damped oscillations

Damping and mechanical energy

In the chapters on simple harmonic motion, we ignored friction, so our model described oscillations that continued forever with the same amplitude. Real oscillating systems, if left alone, will decrease in amplitude and eventually come to rest.

Dissipative forces like friction and drag convert mechanical energy into thermal energy. By the energy principle, with no work done by external forces,

\[\Delta E_{\mathrm{mech}} + \Delta E_{\mathrm{th}} = 0.\]

As thermal energy increases, the mechanical energy decreases. Since $E_{\mathrm{mech}} = \tfrac{1}{2} k_s A^2$, the amplitude of the oscillation gradually decreases. An oscillation that loses energy in this way is called a damped oscillation.

A model of damping: viscous drag

As a simple model, consider a mass moving in a viscous fluid that exerts a drag force proportional to its velocity:

\[F_{d,x} = -b v_x.\]

The constant $b$ depends on the fluid and on the size and shape of the block. The minus sign means the drag force always opposes the motion. With the drag force included, Newton’s second law for the block is

\[-k_s x - b v_x = m a_x.\]

The acceleration is no longer proportional to $-x$, so this is no longer simple harmonic motion. We won’t solve this equation, but we can describe what its solutions look like.

Weakly damped oscillations

When the damping is weak, the block still oscillates, with nearly the same period as without damping. However, the amplitude decreases exponentially over time:

\[\boxed{A(t) = A_0 e^{-\gamma t}.}\]

Here $A_0$ is the initial amplitude, and $\gamma = b/(2m)$ is the damping coefficient, with units of $\mathrm{s^{-1}}$.

A larger damping coefficient makes the amplitude decrease more quickly. A more viscous fluid (larger $b$) or a lighter block (smaller $m$) increases $\gamma$. With strong enough damping, the block returns slowly to equilibrium without oscillating at all. We will only consider weak damping.

The amplitude envelope

The displacement oscillates between the dashed curves $\pm A(t)$, called the envelope. Each time interval $1/\gamma$ reduces the amplitude by the same factor, $e \approx 2.72$.

A graph of displacement x versus time t for a weakly damped oscillator released from rest at x equals A sub 0. The solid blue curve oscillates about ten times, and its peaks get steadily smaller. Two dashed purple curves, labeled plus A of t and minus A of t, are decaying exponentials that start at plus and minus A sub 0 and just touch the peaks and troughs of the oscillation, forming an envelope around it. The vertical axis has ticks at plus A sub 0, A sub 0 over e, and minus A sub 0. Dashed guide lines rise from ticks at 1 over gamma and 2 over gamma on the time axis to the upper envelope, and a dashed horizontal line from the A sub 0 over e tick meets the envelope at 1 over gamma: there the amplitude is A sub 0 over e, about 0.37 A sub 0. At 2 over gamma it is A sub 0 over e squared, about 0.14 A sub 0.

Energy of a damped oscillator

With weak damping, the amplitude changes very little during one cycle, so each cycle looks almost exactly like SHM. The mechanical energy is then approximately $E_{\mathrm{mech}} \approx \tfrac{1}{2} k_s A^2$. Substituting the decreasing amplitude $A(t) = A_0 e^{-\gamma t}$ gives

\[E_{\mathrm{mech}}(t) \approx \tfrac{1}{2} k_s A_0^2 e^{-2\gamma t}.\]

The energy also decreases exponentially, with $2\gamma$ in place of $\gamma$, because the energy depends on the square of the amplitude. When the amplitude has decreased to half its initial value, only a quarter of the mechanical energy remains. The mechanical energy lost by the oscillator becomes thermal energy of the block and the fluid.

Check your understanding: The amplitude of a weakly damped pendulum decreases from $10.0\ \mathrm{cm}$ to $5.0\ \mathrm{cm}$ in $20\ \mathrm{s}$.

  1. What fraction of its initial mechanical energy does the pendulum still have at $t = 20\ \mathrm{s}$?
  2. What is its amplitude at $t = 40\ \mathrm{s}$?
Answer

1. The amplitude has halved, and the mechanical energy is proportional to $A^2$, so $(1/2)^2 = 1/4$ of the initial mechanical energy remains. The other $3/4$ has become thermal energy.

2. Each $20\ \mathrm{s}$ interval multiplies the amplitude by the same factor, $1/2$. So at $t = 40\ \mathrm{s}$ the amplitude is $\tfrac{1}{2}(5.0\ \mathrm{cm}) = 2.5\ \mathrm{cm}$.

Driven oscillations and resonance

Driven oscillations

A real oscillator with a damping force eventually comes to rest. To keep it oscillating, something must add energy to replace the energy lost to damping. An external force that pushes back and forth periodically is called a driving force, and the motion it produces is a driven oscillation.

A familiar example of a driven oscillation is pushing someone on a swing. Another example is a speaker, which drives the air in front of it, producing a sound wave that in turn drives your eardrum.

We will consider a driving force that oscillates sinusoidally with driving angular frequency $\omega_d$. The oscillator also has its own natural angular frequency, the angular frequency at which it oscillates when left alone with no damping. For a mass on a spring, that is

\[\omega_0 = \sqrt{\frac{k_s}{m}}.\]

Transient and steady state

When the driving force first starts, the motion is complicated. The oscillator’s own damped oscillation at its natural frequency adds to a response at the driving frequency. This early stage is the transient.

Damping makes the natural oscillation die out, just as it does for an oscillator with no driving force. The oscillator eventually reaches a steady state, in which it oscillates with a constant amplitude. In the steady state, the oscillator moves at the driving frequency $\omega_d$, not at its natural frequency $\omega_0$. Each cycle, the driving force adds exactly as much energy as damping removes.

The plots below compare two identical oscillators that start at rest at equilibrium. They are driven by sinusoidal forces with the same amplitude but different frequencies. Both graphs use the same vertical scale. The dashed lines indicate the steady-state amplitudes.

Two stacked graphs of displacement x versus time t, from 0 to 12 natural periods T sub 0, drawn to the same vertical scale. Top graph, driven at the natural frequency, omega sub d equals omega sub 0: the blue curve starts at zero and oscillates with an amplitude that grows over the first several cycles, then levels off at a large constant amplitude after about 6 periods, reaching a pair of dashed horizontal lines that mark the steady-state amplitude. The top graph is labeled at resonance. Bottom graph, labeled off resonance, driven at omega sub d equals 1.6 omega sub 0: the curve wobbles irregularly at first, with an amplitude that changes from cycle to cycle but never becomes large, and settles after about 6 periods into a steady oscillation with an amplitude about one eighth as large as in the top graph, again reaching its own pair of dashed lines.

Resonance

The steady-state amplitude depends on the driving frequency and on the damping. It is largest when the driving frequency is close to the natural frequency. This large response is called resonance.

A graph of steady-state amplitude versus driving angular frequency omega sub d, from 0 to 2.5 omega sub 0, for three oscillators that differ only in their damping and are driven by forces of the same size. All three curves start at the same small amplitude at low driving frequency and fall toward zero at high driving frequency. Each has a peak at omega sub d close to omega sub 0, marked by a dashed vertical line. A legend identifies the curves. The solid curve, for weak damping, has a tall, narrow peak about ten times the low-frequency amplitude. The dashed curve, for medium damping, has a peak half as tall and wider. The dotted curve, for strong damping, has a low, broad bump about twice the low-frequency amplitude.

The weaker the damping, the taller and narrower the resonance peak.

Why resonance happens

Think about pushing a child on a swing. If you push once per cycle, always while the swing is moving away from you, every push adds energy. The amplitude builds up until the energy lost to damping each cycle balances the energy you add.

If you push at some other frequency, the swing ends up moving at your frequency, but your pushes no longer stay in step with its motion. For part of every cycle you push against its motion and take energy away, so the steady-state amplitude is smaller than at resonance.

Resonance happens when a driving force stays in step with an oscillator’s velocity, which requires a driving frequency close to the natural frequency. We will meet it again in musical instruments, where strings and air columns respond strongly to particular frequencies.

Example: Finding the resonance

A $0.500\ \mathrm{kg}$ block on a spring with $k_s = 200\ \mathrm{N/m}$ is weakly damped. You drive it with a sinusoidal force, slowly increasing the driving frequency from $1\ \mathrm{Hz}$ to $6\ \mathrm{Hz}$. At what driving frequency is the steady-state amplitude largest?

Solution

The natural angular frequency of the block is

\[\omega_0 = \sqrt{\frac{k_s}{m}} = \sqrt{\frac{200\ \mathrm{N/m}}{0.500\ \mathrm{kg}}} = 20.0\ \mathrm{rad/s},\]

so its natural frequency is

\[f_0 = \frac{\omega_0}{2\pi} = 3.18\ \mathrm{Hz}.\]

With weak damping, the resonance peak is close to the natural frequency, so the steady-state amplitude is largest for a driving frequency close to $3.18\ \mathrm{Hz}$.