Damped and Forced Oscillations

Lessons

Senior 5 · Physics

Damped and Forced Oscillations

3 lessons

Lesson 3 of 3

Forced Oscillations and Resonance

25 min

Pushing at the right moment

Anyone who has pushed a child on a swing knows the trick: small pushes, timed to arrive once per swing just as the swing starts to move away from you, build up a large motion. Push at random times and almost nothing happens. The same physics makes a washing machine shake violently for a few seconds as its drum speeds up, makes a bus window buzz at one particular engine speed, and once closed a brand-new footbridge in London two days after it opened. In each case an outside force is driving an oscillator, and the response depends dramatically on how the driving frequency compares with the natural frequency.

Free and forced oscillations

A system displaced and then left alone oscillates at its natural frequency $f_0 = \frac{\omega_0}{2\pi}$ ; this is a free oscillation. If instead a periodic force $F = F_0\cos\omega t$ keeps acting on it, the system performs a forced (driven) oscillation. Newton's second law now reads

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

At first the motion is messy, a mixture of the natural oscillation and the driven one. The natural part dies away through damping (the transient), and the system settles into a steady state: it oscillates at the driving frequency $\omega$ , not at its own, with constant amplitude

A(\omega) = \frac{F_0/m}{\sqrt{\left(\omega_0^2 - \omega^2\right)^2 + \left(2\gamma\omega\right)^2}}, \qquad \gamma = \frac{b}{2m}

(You can check it by substituting $x = A\cos(\omega t - \delta)$ ; the working is in the OpenStax section listed in the credits.) Read the formula at three frequencies:

Very low $\omega$ : $A \to \frac{F_0}{m\omega_0^2} = \frac{F_0}{k}$ , the static deflection. The system simply follows the slowly changing force, in phase with it.
$\omega$ close to $\omega_0$ : the first term under the root vanishes and only damping limits the amplitude: $A \approx \frac{F_0}{2m\gamma\omega_0} = \frac{F_0}{b\omega_0}$ . This is resonance.
Very high $\omega$ : $A \to \frac{F_0}{m\omega^2}$ , tiny. The mass's inertia cannot follow the rapid force, and it moves in antiphase with it.

The lighter the damping, the taller and narrower the peak, and the closer it lies to $\omega_0$ (the peak is at $\sqrt{\omega_0^2 - 2\gamma^2}$ ). The height of the peak, as a multiple of the static deflection, is about $Q$ from Lesson 2. Heavy damping flattens the curve until there is hardly any peak at all.

Phase

The displacement lags behind the driving force by an angle $\delta$ that grows from 0 (at low frequency) through $\frac{\pi}{2}$ (exactly at $\omega_0$ ) to nearly $π \pi$ (at high frequency). At $\omega_0$ the velocity is in phase with the force, which is exactly the swing-pusher's trick: push while the swing moves away from you.

Definition

Resonance is the large increase in the amplitude of a forced oscillation that occurs when the driving frequency is equal (or very close) to the natural frequency of the driven system. A forced oscillation is one maintained by a periodic external force; in the steady state it has the frequency of the driver. Example: a child's swing of natural period 3.0 s is pushed once every 3.0 s; each push adds energy and the amplitude grows until the energy lost per cycle equals the energy added.

Worked example 1: amplitude below, at and above resonance

A 2.0 kg mass on a spring of stiffness $800\,\mathrm{N\,m^{-1}}$ , with damping constant $4.0\,\mathrm{kg\,s^{-1}}$ , is driven by a force of amplitude 10 N. Find the steady amplitude when $\omega = 10\,\mathrm{rad\,s^{-1}}$ , $\omega = \omega_0$ and $\omega = 30\,\mathrm{rad\,s^{-1}}$ , and compare with the static deflection.

Solution

$\omega_0 = \sqrt{\frac{800}{2.0}} = 20\,\mathrm{rad\,s^{-1}}$ , $\gamma = \frac{4.0}{4.0} = 1.0\,\mathrm{s^{-1}}$ , $\frac{F_0}{m} = 5.0\,\mathrm{N\,kg^{-1}}$ . Static deflection $\frac{F_0}{k} = 0.0125\,\mathrm{m}$ .

$\omega = 10$ : $A = \frac{5.0}{\sqrt{(400 - 100)^2 + (2 \times 1.0 \times 10)^2}} = \frac{5.0}{300.7} = 0.0166\,\mathrm{m}$ .

$\omega = 20$ : $A = \frac{5.0}{\sqrt{0 + (2 \times 1.0 \times 20)^2}} = \frac{5.0}{40} = 0.125\,\mathrm{m}$ , ten times the static deflection ( $Q = \frac{\omega_0}{2\gamma} = 10$ ).

$\omega = 30$ : $A = \frac{5.0}{\sqrt{(400 - 900)^2 + 60^2}} = \frac{5.0}{503.6} = 0.0099\,\mathrm{m}$ , smaller than the static deflection.

Energy in forced oscillations

In the steady state the amplitude is constant, so the oscillator's stored energy is constant. Every cycle, the driver does work on the system, and the damping force converts exactly the same amount into internal energy:

\text{mean power from the driver} = \text{mean power dissipated} = \tfrac{1}{2}b\,(\omega A)^2

( $\omega A$ is the peak speed and the mean of $v^{2}$ over a cycle is $\frac{1}{2}(\omega A)^2$ .) Away from resonance, the force is partly in step with the velocity and partly against it: during some of each cycle the driver does negative work and takes energy back, so little energy is stored. At resonance the force is in phase with the velocity all the time, the driver always does positive work, and the power transfer is greatest. The amplitude then grows until the energy lost per cycle, which rises as $A^{2}$ , catches up with the energy supplied.

Worked example 2: the energy balance at resonance

For the system of Worked example 1 driven at $\omega_0$ , find the mean power dissipated and show that it equals the mean power supplied by the driver.

Solution

Peak speed $v_0 = \omega A = 20 \times 0.125 = 2.5\,\mathrm{m\,s^{-1}}$ . Dissipated: $\frac{1}{2}bv_0^2 = \frac{1}{2} \times 4.0 \times 2.5^2 = 12.5\,\mathrm{W}$ .

Supplied: at resonance $F$ and $v$ are in phase, so $P = F_0\cos\omega t \times v_0\cos\omega t$ , whose mean is $\frac{1}{2}F_0v_0 = \frac{1}{2} \times 10 \times 2.5 = 12.5\,\mathrm{W}$ . The two agree, as they must when the amplitude is steady.

Experiment 2.2: Barton's pendulums

Aim: to observe how the amplitude and phase of a driven oscillator depend on the relation between its natural frequency and the driving frequency.

Apparatus: a string stretched between two clamp stands; a "driver" pendulum with a heavy bob (a 200 g mass); six light pendulums made of paper cones with a little plasticine, of lengths from about 0.3 m to 1.0 m, one of them the same length as the driver.

Method and observations: Pull the driver aside, perpendicular to the string, and release it. Its swing makes the string twitch, which drives all the light pendulums at the driver's frequency. After about a minute of transient motion:

the pendulum of the same length as the driver swings with the largest amplitude, and about a quarter of a cycle ( $\frac{\pi}{2}$ ) behind the driver;
shorter pendulums (higher natural frequency) swing with small amplitude, nearly in phase with the driver;
longer pendulums (lower natural frequency) swing with small amplitude, nearly in antiphase.

Now add a ring of card to each cone to increase the air damping, and repeat: all amplitudes fall, and the peak at the matching length becomes less pronounced, exactly as the resonance curves predict.

Warning

Safety. Clamp both stands firmly to the bench, or weight their bases, since the stretched string pulls them inwards. Keep the heavy driver bob's swing clear of people's faces.

Resonance at work, and resonance as a danger

Useful resonance	What is driven, and by what
Tuning a radio	An electrical circuit, adjusted until its natural frequency equals that of one station's signal, responds strongly to that station only.
Musical instruments	The air in a flute or the body of a guitar or an inanga resonates with the vibrating lips, reed or strings, making the note loud.
Quartz clocks	A circuit drives the crystal of Unit 1 at its resonant frequency; its very high $Q$ keeps the frequency precise.
Magnetic resonance imaging (MRI)	Hydrogen nuclei in the body absorb radio waves at their resonant frequency in a strong magnetic field.

Resonance can also do damage. Machines are mounted on rubber blocks so that their vibration frequency is far from the natural frequency of the floor. Bridges and tall buildings are designed so that the forces of wind, traffic, footsteps and earthquakes cannot build up large oscillations:

On 12 April 1831 the Broughton Suspension Bridge near Manchester, England, collapsed while soldiers marched across it in step, reportedly because their steps drove the bridge at resonance. The British Army then ordered troops to "break step" when crossing bridges.
London's Millennium Footbridge opened on 10 June 2000 and swayed sideways so much under crowds that it was closed within days. Walkers unconsciously fell into step with the sway, driving it at its natural frequency. Dampers were fitted, and it reopened in February 2002.
Taipei 101, a tower in Taiwan, hangs a steel ball of about 660 tonnes near its top. This tuned mass damper swings out of phase with the building and, through its damping, takes energy from the building's sway in strong winds.

Important

Evaluate a famous example. Many books say the Tacoma Narrows Bridge in the USA collapsed on 7 November 1940 because the wind drove it at its natural frequency. Engineers who studied it found something different. The wind was steady, not a periodic force at the bridge's frequency. Instead, once the deck began to twist, its own motion changed how the air flowed round it, and the air then pushed on the deck in step with the twisting. This self-excited motion, called aeroelastic flutter, fed in more energy each cycle than damping removed. It is related to resonance (the bridge oscillated at one of its natural frequencies), but the driving was produced by the motion itself. Modern bridge decks are shaped and tested in wind tunnels to prevent it.

Worked example 3: the shaking washing machine

A washing machine on its springs has a natural frequency of 4.0 Hz (an illustrative value). Its drum speeds up from rest to 1200 revolutions per minute, and an uneven load of clothes acts as a rotating driving force at the drum's frequency. At what drum speed does the machine shake most, and why does it calm down at full speed?

Solution

Resonance occurs when the driving frequency equals 4.0 Hz, that is $4.0 \times 60 = 240$ revolutions per minute. At full speed the driving frequency is $\frac{1200}{60} = 20\,\mathrm{Hz}$ , five times the natural frequency; far above resonance the amplitude is small (it falls as $\frac{1}{\omega^2}$ ). The machine shakes hard only while the drum passes through about 240 rpm; good machines also add dampers to keep that peak low.

Warning

Common mistakes. (1) Saying a forced oscillator vibrates at its natural frequency: in the steady state it vibrates at the driving frequency. (2) Saying resonance makes the amplitude infinite: damping always limits it. (3) Forgetting that damping both lowers the peak and moves it slightly below $\omega_0$ . (4) Calling every large vibration "resonance" without checking that a periodic force matches a natural frequency; the Tacoma Narrows collapse is the classic example.

Tip

For any resonance question, name three things: the oscillator (what has the natural frequency), the driver (what provides the periodic force) and the damping (what removes energy). Then the answer usually writes itself.

Why the Tacoma Narrows Bridge Collapsed (Practical Engineering). Watch for the difference between ordinary resonance and the flutter that actually destroyed the bridge, and how bridge decks are now designed.

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Practice

Distinguish between free and forced oscillations, and state the frequency of a forced oscillation in its steady state.
Sketch amplitude–frequency curves for a lightly and a heavily damped system driven by the same force. Label the natural frequency.
A mass–spring system has $\omega_0 = 15\,\mathrm{rad\,s^{-1}}$ , $\gamma = 0.50\,\mathrm{s^{-1}}$ and $\frac{F_0}{m} = 3.0\,\mathrm{N\,kg^{-1}}$ . Find its amplitude at resonance and when driven at $5.0\,\mathrm{rad\,s^{-1}}$ .
For the system in question 3, the mass is 0.40 kg. Find the mean power dissipated at resonance.
In Barton's pendulums, describe the phase of (a) a short pendulum, (b) the matching pendulum, (c) a long pendulum, relative to the driver.
A minibus window buzzes loudly only when the engine runs at about 1800 revolutions per minute. Explain, and estimate the window's natural frequency if it is driven once per revolution.
Explain why soldiers are told to break step when crossing a bridge, and why this is not needed on a solid stone road.
A student writes: "The Tacoma Narrows Bridge fell because gusts of wind pushed it at its natural frequency." Evaluate this statement.

Show answersHide answers

A free oscillation happens when a system is displaced and released; it vibrates at its natural frequency. A forced oscillation is maintained by a periodic external force; in the steady state it vibrates at the driving frequency.
Both start at the static deflection and fall to zero at high frequency. The lightly damped curve has a tall, narrow peak at (just below) $f_{0}$ ; the heavily damped curve has a low, broad peak further below $f_{0}$ .
Resonance: $A = \frac{3.0}{2 \times 0.50 \times 15} = 0.20\,\mathrm{m}$ . At 5.0: $A = \frac{3.0}{\sqrt{(225 - 25)^2 + (5.0)^2}} = \frac{3.0}{200.1} = 0.015\,\mathrm{m}$ .
$b = 2m\gamma = 0.40\,\mathrm{kg\,s^{-1}}$ ; $v_0 = 15 \times 0.20 = 3.0\,\mathrm{m\,s^{-1}}$ ; $P = \frac{1}{2} \times 0.40 \times 3.0^2 = 1.8\,\mathrm{W}$ .
(a) nearly in phase; (b) a quarter cycle ( $\frac{\pi}{2}$ ) behind; (c) nearly in antiphase ( $π \pi$ behind).
The engine's vibration drives the window; at 1800 rpm the driving frequency matches the window's natural frequency and it resonates. $f_0 \approx \frac{1800}{60} = 30\,\mathrm{Hz}$ .
Marching in step applies a periodic force at the step frequency; if this matches a natural frequency of a light, flexible bridge, the amplitude can build up dangerously. Breaking step makes the force irregular. A road on solid ground has no low natural frequency that footsteps could drive.
It is not accurate. The wind was steady, not periodic at the bridge's frequency. The collapse was caused by aeroelastic flutter: the deck's own twisting made the airflow push it in step with its motion, feeding in energy faster than damping removed it.

Key points

Steady forced oscillations have the driving frequency and amplitude $A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}}$ .
Resonance: amplitude is greatest when the driving frequency is close to the natural frequency; at the peak, $A \approx \frac{F_0}{b\omega_0}$ , about $Q$ times the static deflection.
Lighter damping gives a taller, sharper peak; the displacement lags the force by $\frac{\pi}{2}$ at $\omega_0$ .
In the steady state, power in from the driver equals power dissipated, $\frac{1}{2}b(\omega A)^2$ .

1Pushing at the right moment

2Free and forced oscillations

Phase

3Energy in forced oscillations

4Experiment 2.2: Barton's pendulums

5Resonance at work, and resonance as a danger

Pushing at the right moment

Free and forced oscillations

Energy in forced oscillations

Experiment 2.2: Barton's pendulums

Resonance at work, and resonance as a danger