Damped and Forced Oscillations

Lessons

Senior 5 · Physics

Damped and Forced Oscillations

3 lessons

Lesson 2 of 3

Energy in Damped Oscillations

25 min

Louder, but shorter

Strike a tuning fork and hold it in the air: you hear a faint note that lasts for many seconds. Strike it again and press its stem onto a wooden table: the note is suddenly much louder, but it dies away much sooner. The table is not adding energy; it is helping the fork give its energy away faster, as sound spreading from a large vibrating surface. Where the energy of a damped oscillator goes, and how fast it leaves, decides how long a bell rings, how sharply a radio tunes and how quickly a bridge stops swaying.

How the energy decays

From Unit 1, an oscillator's total energy is $E = \frac{1}{2}kA^2$ . With light damping the amplitude decays as $A = A_0e^{-\gamma t}$ , so

E = \tfrac{1}{2}kA_0^2e^{-2\gamma t} = E_0e^{-2\gamma t} = E_0e^{-\frac{b}{m}t}

The energy decays exponentially at twice the rate of the amplitude. When the amplitude has halved, only a quarter of the energy is left.

The rate of energy loss

Differentiate the mechanical energy $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$ with respect to time and use the equation of motion $m a = - k x - b v$ :

\frac{dE}{dt} = mv\frac{dv}{dt} + kx\frac{dx}{dt} = v(ma + kx) = v(-bv) = -bv^2

So the oscillator loses energy at the rate $b v^{2}$ : fastest where it moves fastest (through equilibrium) and not at all at the extremes, where it is momentarily at rest. This lost mechanical energy does not disappear; the damping force does work on the surroundings, and the energy ends up as internal energy (a tiny rise in temperature of the air, the oil or the spring) and as sound.

Worked example 1: amplitude and energy

A 0.20 kg mass on a spring of stiffness $20\,\mathrm{N\,m^{-1}}$ starts with amplitude 10 cm. Some time later its amplitude is 5.0 cm. How much mechanical energy has been converted to other forms?

Solution

$E_0 = \frac{1}{2} \times 20 \times 0.10^2 = 0.10\,\mathrm{J}$ and $E = \frac{1}{2} \times 20 \times 0.050^2 = 0.025\,\mathrm{J}$ . The energy converted is $0.10 - 0.025 = 0.075\,\mathrm{J}$ : 75% of the energy has gone while the amplitude has fallen by only half.

The quality factor

How good is an oscillator at keeping its energy? The fraction lost in one period $T$ is

\frac{\Delta E}{E} = 1 - e^{-2\gamma T} \approx 2\gamma T = \frac{4\pi\gamma}{\omega_0} \quad\text{(for light damping)}

Engineers turn this round and define the quality factor $Q$ :

Q = 2\pi \times \frac{\text{energy stored}}{\text{energy lost per cycle}} \approx \frac{\omega_0}{2\gamma} = \frac{m\omega_0}{b}

A high $Q$ means light damping: a tuning fork, a bell or the quartz crystal of Unit 1 rings for many cycles. A low $Q$ means heavy damping: a car suspension, designed to stop bouncing within about one cycle, has $Q$ of order 1. A useful rule: the amplitude falls to $\frac{1}{e}$ (37%) of its starting value in about $\frac{Q}{\pi}$ cycles.

Definition

The quality factor $Q$ of an oscillator is $2\pi$ times the ratio of the energy stored to the energy lost per cycle. For light damping $Q \approx \frac{\omega_0}{2\gamma} = \frac{m\omega_0}{b}$ . Example: the spring in Worked example 1 of Lesson 1 ( $\omega_0 = 10\,\mathrm{rad\,s^{-1}}$ , $\gamma = 0.40\,\mathrm{s^{-1}}$ ) has $Q \approx \frac{10}{0.80} = 12.5$ .

Experiment 2.1: measuring the damping of a spring

Aim: to show that the amplitude of a damped oscillator decays exponentially, and to find $\gamma$ , $b$ and $Q$ .

Apparatus: a spring, a 400 g slotted mass on a hanger, a light card about 10 cm square fixed under the mass, a clamp stand, a metre rule clamped vertically beside the mass, a stopwatch (or a phone video, which you can step through frame by frame).

Method:

Time 20 oscillations to find the period $T$ .
Pull the mass down about 10 cm and release it. Start the stopwatch.
Every 10 oscillations (every 8 s here), read the highest point of the pointer against the rule and subtract the rest position to get the amplitude $A$ .
Plot $\ln A$ against $t$ . Since $A = A_0e^{-\gamma t}$ , $\ln A = \ln A_0 - \gamma t$ : a straight line of gradient $-\gamma$ .

Illustrative results ( $m = 0.400\,\mathrm{kg}$ , $T = 0.80\,\mathrm{s}$ ):

$t$ / s	$A$ / cm	$\ln(A / \text{cm})$
0	10.0	2.30
8	7.4	2.00
16	5.5	1.70
24	4.1	1.41
32	3.0	1.10
40	2.2	0.79

The points lie on a straight line, which confirms exponential decay. Its gradient is $-0.0378\,\mathrm{s^{-1}}$ , so $\gamma = 0.0378\,\mathrm{s^{-1}}$ . Then $b = 2m\gamma = 2 \times 0.400 \times 0.0378 = 0.030\,\mathrm{kg\,s^{-1}}$ . With $\omega_0 = \frac{2\pi}{0.80} = 7.85\,\mathrm{rad\,s^{-1}}$ , $Q \approx \frac{\omega_0}{2\gamma} = 104$ . Repeat with a larger card: $\gamma$ rises and $Q$ falls.

Warning

Safety. Wear eye protection when loading springs: an overloaded spring can snap or fly off its support. Do not stretch the spring beyond its elastic limit, keep the masses low over the bench, and keep your feet clear of the hanging load.

Worked example 2: energy lost per cycle

Using the results of Experiment 2.1, find (a) the fraction of the energy lost in one cycle, (b) the time for the energy to halve.

Solution

(a) $\frac{\Delta E}{E} = 1 - e^{-2\gamma T} = 1 - e^{-2 \times 0.0378 \times 0.80} = 1 - e^{-0.0604} = 0.059$ : about 6% per cycle. (Check: $\frac{2\pi}{Q} = \frac{2\pi}{104} = 0.060$ .)

(b) $e^{-2\gamma t} = \frac{1}{2}$ gives $t = \frac{\ln 2}{2\gamma} = \frac{0.693}{0.0755} = 9.2\,\mathrm{s}$ , about 11 oscillations. The amplitude takes twice as long, 18 s, to halve.

Warning

Common mistakes. (1) Using the amplitude's decay rate for the energy: energy decays as $e^{-2\gamma t}$ , twice as fast. (2) Saying the damped energy is "destroyed": it is transferred to internal energy and sound. (3) Plotting $A$ against $t$ and trying to read $\gamma$ from a curve; plot $\ln A$ to get a straight line. (4) Mixing $b$ , $\gamma = \frac{b}{2m}$ and $Q = \frac{m\omega_0}{b}$ : write down which one the question asks for.

Tip

"Energy goes as amplitude squared" settles most damping questions in one line: amplitude down to 0.9 means energy down to ${0.9}^{2} = 0.81$ , so 19% of the energy has gone.

Damping and resonance: A-level Physics (Science Shorts). Watch for how the amplitude envelope shrinks under light, critical and heavy damping, and how damping changes the resonance peak you will meet in the next lesson.

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Practice

The amplitude of a damped oscillator falls from 6.0 cm to 3.0 cm. What fraction of its energy remains?
Show that the rate of loss of mechanical energy of a damped oscillator is $b v^{2}$ . Where in the cycle is it greatest?
An oscillator loses 2.0% of its amplitude each cycle. What percentage of its energy does it lose each cycle?
A 0.50 kg mass on a spring has $\omega_0 = 12\,\mathrm{rad\,s^{-1}}$ and damping constant $b = 0.060\,\mathrm{kg\,s^{-1}}$ . Find $\gamma$ and $Q$ .
For the oscillator in question 4, find the time for its energy to fall to 10% of its starting value.
In Experiment 2.1 a student plots $A$ against $t$ and gets a curve. Explain why a graph of $\ln A$ against $t$ is better, and what its gradient gives.
Explain, in terms of energy and $Q$ , why a tuning fork pressed on a table sounds louder but for a shorter time.
A 2.0 kg pendulum bob swinging with amplitude 0.30 m in a large hall has $Q = 400$ and period 2.0 s. Estimate the mean power needed to keep it swinging with constant amplitude. (Take $E = \frac{1}{2}m\omega_0^2A^2$ .)

Show answersHide answers

$\left(\frac{3.0}{6.0}\right)^2 = \frac{1}{4}$ : a quarter.
$\frac{dE}{dt} = v(ma + kx)$ and $m a + k x = - b v$ , so $\frac{dE}{dt} = -bv^2$ . Greatest where $v$ is greatest, at equilibrium.
Energy left ${0.98}^{2} = 0.960$ : about 4.0% lost per cycle.
$\gamma = \frac{0.060}{1.0} = 0.060\,\mathrm{s^{-1}}$ ; $Q = \frac{12}{0.12} = 100$ .
$e^{-2\gamma t} = 0.10$ , so $t = \frac{\ln 10}{0.12} = 19\,\mathrm{s}$ .
A straight line is easy to recognise and to fit. If $\ln A$ against $t$ is straight, the decay is exponential, and the gradient is $-\gamma$ .
The large table surface radiates sound far more effectively, so energy leaves the fork faster: the damping is larger and $Q$ smaller. A larger rate of energy loss means more sound power (louder), and the stored energy is used up sooner (shorter).
$\omega_0 = \pi\,\mathrm{rad\,s^{-1}}$ ; $E = \frac{1}{2} \times 2.0 \times \pi^2 \times 0.30^2 = 0.888\,\mathrm{J}$ . Energy lost per cycle $= \frac{2\pi E}{Q} = \frac{2\pi \times 0.888}{400} = 0.0140\,\mathrm{J}$ , every 2.0 s: about $7.0 \times 10^{-3}\,\mathrm{W}$ (7 mW).

Key points

$E = E_0e^{-2\gamma t}$ : energy decays twice as fast as amplitude; $E \propto A^2$ .
Rate of energy loss $= b v^{2}$ ; the energy becomes internal energy of the surroundings and sound.
$Q = 2\pi \frac{E}{\Delta E_{\text{cycle}}} \approx \frac{\omega_0}{2\gamma}$ ; high $Q$ means a long ring-down.
Plot $\ln A$ against $t$ : a straight line of gradient $-\gamma$ .

Sources and credits

15.5 Damped Oscillations, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Damping & Resonance - A-level Physics, Science Shorts, embedded from YouTube.

The experiment and its illustrative results, the three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Louder, but shorter

2How the energy decays

The rate of energy loss

3The quality factor

4Experiment 2.1: measuring the damping of a spring

Louder, but shorter

How the energy decays

The quality factor

Experiment 2.1: measuring the damping of a spring