An old bus idles at a traffic signal, and one of its windows rattles loudly. The other windows are quiet. When the light turns green and the driver presses the accelerator, the rattle stops. The engine shakes the whole bus, yet only one window answers, and only at one engine speed.
Free and forced
Pull a block on a spring aside and let go. It oscillates at its own natural frequency \(\omega_0 = \sqrt{k/m}\), with an amplitude set by the pull. This is a free oscillation, and with damping it dies away. Now let an outside agent push the block to and fro with a force \(F = F_0\sin\omega t\). The block is forced to move, and its motion is a forced oscillation. The angular frequency \(\omega\) of the push is the driving frequency. It is chosen by the agent and has nothing to do with the spring.
With damping and the push together, Newton's second law becomes \[ m\dfrac{d^2x}{dt^2} + b\dfrac{dx}{dt} + kx = F_0\sin\omega t. \] Its solution has two parts. One part is the damped free oscillation of the last lesson, at \(\omega'\), which decays as \(e^{-bt/2m}\). The other is an oscillation at the driving frequency \(\omega\), with a steady amplitude. After a few multiples of \(2m/b\) only the second part is left. This lasting motion is called the steady state.
Try \(0.5\,\text{Hz}\) in the figure above. The block wobbles unevenly for a few seconds while the free part dies. Then it settles into a steady motion at \(0.5\,\text{Hz}\), in step with the driver, with a swing of only \(0.53\,\text{cm}\). At \(1.6\,\text{Hz}\) it settles at \(0.26\,\text{cm}\), less than the driver's own \(0.4\,\text{cm}\), and moves almost opposite to the driver. At \(1.0\,\text{Hz}\) the swing grows for about ten seconds and settles near \(4.2\,\text{cm}\), about ten times the driver's motion.
A block on a spring has natural frequency \(2\,\text{Hz}\). It is driven by a periodic force of frequency \(3\,\text{Hz}\), and it has some damping. At what frequency does it oscillate once the steady state is reached?
How large is the steady swing?
Substituting \(x = A\sin(\omega t - \theta)\) into the driven equation gives the steady amplitude \(A\) and the lag \(\theta\) of the motion behind the force. The algebra is long, but the result is short.
Read the formula at three driving frequencies. For a very slow push, \(\omega \to 0\), it gives \(A = F_0/(m\omega_0^2) = F_0/k\). The spring simply stretches with the force. For a very fast push, \(\omega^2\) dominates the bottom, and \(A \approx F_0/(m\omega^2)\). This tends to zero, because the block's inertia cannot follow. At \(\omega = \omega_0\) the first bracket is zero. Only the damping term is left, and \(A = F_0/(b\omega_0)\).
Resonance is the large response of an oscillator driven at a frequency close to its natural frequency. For light damping the steady amplitude is greatest there, and a small force produces a large oscillation.
Why is the natural frequency special? At resonance the steady motion lags the force by exactly a quarter cycle, \(\theta = \pi/2\). The velocity then peaks exactly when the force peaks, so the force always pushes in the direction of motion. Every push does positive work, like pushes on a swing given at just the right moment. The amplitude grows until the damping loses energy as fast as the force supplies it.
A \(2\,\text{kg}\) block on a spring of \(200\,\text{N/m}\), with damping constant \(4\,\text{kg/s}\), is driven by a force of amplitude \(2\,\text{N}\). Find the steady amplitude for a very slow push, for \(\omega = \omega_0\) and for \(\omega = 20\,\text{rad/s}\).
\(\omega_0 = \sqrt{200/2} = 10\,\text{rad/s}\), \(F_0/m = 1\,\text{N/kg}\) and \(b/m = 2\,\text{s}^{-1}\).
Slow push: \(A = F_0/k\). At resonance: \(A = F_0/(b\omega_0)\). At \(20\,\text{rad/s}\): \(A = (F_0/m)/\sqrt{(\omega_0^2 - \omega^2)^2 + (b\omega/m)^2}\).
Slow: \(2/200 = 0.01\,\text{m} = 1\,\text{cm}\). Resonance: \(2/(4 \times 10) = 0.05\,\text{m} = 5\,\text{cm}\). At \(20\,\text{rad/s}\): \(1/\sqrt{(100 - 400)^2 + 40^2} = 1/\sqrt{91\,600} \approx 0.0033\,\text{m}\).
The same \(2\,\text{N}\) force gives \(1\,\text{cm}\) pushed slowly, \(5\,\text{cm}\) at resonance and only \(0.33\,\text{cm}\) at twice the natural frequency.
A lightly damped oscillator is driven at its natural frequency. What happens to the amplitude of its steady oscillation?
A \(0.4\,\text{kg}\) block hangs on a spring of \(160\,\text{N/m}\). A motor shakes the support. At what frequency of the motor, in hertz, does the block resonate?
A playground swing hangs on chains \(2.5\,\text{m}\) long. You push once in every swing. Estimate the time between your pushes, in seconds. Take \(g = 10\,\text{m/s}^2\).
Damping shapes the peak
The height of the peak is about \(F_0/(b\omega_0)\), so it is set by the damping alone. Halve \(b\), and the peak doubles. Away from the peak, the term \((\omega_0^2 - \omega^2)^2\) is much larger than the damping term, so \(b\) hardly matters there. So less damping makes the peak both taller and narrower: the oscillator answers strongly to a small range of frequencies and weakly to all others.
Move the slider in the figure above. At \(b = 1\,\text{kg/s}\) the peak is \(6.4\,\text{cm}\) at \(0.99\,\text{Hz}\), more than six times the static stretch of about \(1\,\text{cm}\). At \(b = 2\,\text{kg/s}\) it drops to \(3.2\,\text{cm}\), at \(0.97\,\text{Hz}\). At \(b = 6\,\text{kg/s}\) only a gentle bump of \(1.2\,\text{cm}\) remains, at \(0.74\,\text{Hz}\). The left end of every curve stays near \(1\,\text{cm}\), and the right end falls towards zero whatever the damping.
The top of the amplitude curve is not exactly at \(\omega_0\). Minimising the bottom of the formula gives the peak at \(\omega_r = \sqrt{\omega_0^2 - b^2/2m^2}\), a little below \(\omega_0\). For light damping the difference is tiny, which is why resonance is usually said to occur at the natural frequency. The energy the oscillator absorbs, as distinct from its amplitude, peaks exactly at \(\omega_0\).
An oscillator has natural angular frequency \(5\,\text{rad/s}\) and damping constant \(2\,\text{kg/s}\). It is driven at \(5\,\text{rad/s}\) by a force of amplitude \(0.5\,\text{N}\). What is its steady amplitude, in cm?
The damping of a driven oscillator is reduced, with everything else unchanged. Which statements are true?
Resonance, useful and dangerous
Every structure has natural frequencies: a bridge, a building, a wing, a wine glass. A periodic force at one of those frequencies can build a large motion from small pushes. So engineers check that wind, traffic and footsteps do not drive a bridge near its natural frequency. When the Millennium Bridge in London opened in 2000, walkers fell into step with its gentle sideways sway, and it swayed more. It was closed and fitted with dampers.
The same effect is useful wherever one frequency must be picked out from many. A radio receiver tunes its circuit's natural frequency to that of one station, and the circuit resonates with that station alone. The body of a veena or a tabla resonates and strengthens certain notes. In an MRI scanner, hydrogen nuclei in the body absorb radio waves strongly only at their own natural frequency. That absorption builds the image. Tall buildings carry a heavy tuned damper, a mass on springs or a pendulum tuned to the building's frequency. It swings against the sway caused by wind or an earthquake and drains its energy.
Why are soldiers told to break step when they march across a bridge?
About seventy soldiers were returning to barracks across the Broughton Suspension Bridge near Manchester, marching four abreast and in step. The bridge began to sway up and down in time with their feet, more with every step. A bolt broke, and one end of the bridge fell into the river. Several soldiers were hurt and none died. After this, the British Army ordered troops to break step when crossing a bridge.
A sitar has, besides its main playing strings, a row of thinner strings underneath, often eleven to thirteen of them. They are never plucked. Each is tuned to a note of the raga being played. When the player sounds a note that matches one of them, that string is driven at its natural frequency through the bridge. It resonates and rings on, giving the sitar its shimmering sound.
Back at the traffic signal, the idling engine shakes the bus at one steady frequency. One window, held loosely in its frame, has a natural frequency close to that engine speed, so it resonates and rattles. The other windows have different natural frequencies and barely respond. When the driver presses the accelerator, the driving frequency rises past the window's natural frequency, and the rattle stops.
- Saying a forced oscillator moves at its natural frequency.In the steady state it moves at the driving frequency. The natural frequency decides only how large the motion is.
- Thinking resonance makes the amplitude infinite.Damping always limits it, to about \(F_0/(b\omega_0)\). Only a frictionless oscillator would grow without limit.
- Thinking that more damping always lowers the amplitude by the same fraction.Damping matters mostly near resonance. Far from it, the amplitude is set by the spring or by the inertia.