Simple Harmonic Motion

Lessons

Senior 5 · Physics

Simple Harmonic Motion

4 lessons

Lesson 4 of 4

Adding Simple Harmonic Motions

20 min

Silence made from sound

Noise-cancelling headphones have a tiny microphone that listens to the noise outside. A circuit then drives the earphone with a second vibration, the "anti-noise", at the same frequency but turned half a cycle out of step. Your eardrum is pushed by both at once, and its displacement is the sum of the two. When the sum is close to zero you hear close to nothing. To understand how good the cancellation can be, you need to add two simple harmonic motions of the same frequency and find the amplitude and phase of the result.

The principle of superposition

When a body is subjected to two simple harmonic motions along the same line at the same time, its displacement at every instant is the sum of the displacements that each motion alone would produce:

x = x_{1} + x_{2}

Take two motions with the same angular frequency $\omega$ but different amplitudes and a phase difference $\delta$ :

x_1 = A_1\sin\omega t, \qquad x_2 = A_2\sin(\omega t + \delta)

Deriving the resultant

Expand $\sin(\omega t + \delta) = \sin\omega t\cos\delta + \cos\omega t\sin\delta$ and collect terms:

x = (A_1 + A_2\cos\delta)\sin\omega t + (A_2\sin\delta)\cos\omega t

We want this in the form $x = A\sin(\omega t + \phi) = (A\cos\phi)\sin\omega t + (A\sin\phi)\cos\omega t$ . Comparing coefficients:

A\cos\phi = A_1 + A_2\cos\delta, \qquad A\sin\phi = A_2\sin\delta

Squaring and adding, then dividing:

A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\delta}, \qquad \tan\phi = \frac{A_2\sin\delta}{A_1 + A_2\cos\delta}

The result is itself a simple harmonic motion, with the same frequency, a new amplitude $A$ and a new phase constant $\phi$ .

Definition

The principle of superposition states that when two or more oscillations act on the same body along the same line, the resultant displacement is the algebraic sum of the individual displacements. Two SHMs of the same frequency combine into one SHM of that frequency. Example: $x_1 = 4.0\sin\omega t$ and $x_2 = 4.0\sin(\omega t + \pi)$ give $x = 0$ at every instant.

Adding with phasors

The same result follows from the reference circle of Lesson 1. Represent each motion by a phasor: an arrow whose length is the amplitude and whose angle from a reference line is the phase. Both phasors turn at the same $\omega$ , so the angle between them stays $\delta$ and the shape of the diagram never changes. The resultant motion is represented by the vector sum of the two phasors, and the cosine rule on the triangle gives exactly the formula for $A$ .

Three special cases

Phase difference $\delta$	Resultant amplitude	Name
$0$ (in phase)	$A_{1} + A_{2}$ , the greatest possible	constructive
$\frac{\pi}{2}$ (quadrature)	$\sqrt{A_1^2 + A_2^2}$	phasors at right angles
$π \pi$ (antiphase)	$∣ A_{1} - A_{2} ∣$ , the least possible	destructive

So, whatever the phase, $|A_1 - A_2| \le A \le A_1 + A_2$ . Complete cancellation needs both equal amplitudes and a phase difference of exactly $π \pi$ .

If the two frequencies are different, the angle between the phasors keeps changing and the sum is no longer SHM: its amplitude rises and falls slowly. You will meet this effect, called beats, with sound waves.

Worked example 1: two motions in quadrature

Find the resultant of $x_1 = 4.0\sin\omega t$ and $x_2 = 3.0\sin\left(\omega t + \frac{\pi}{2}\right)$ , in centimetres.

Solution

With $\delta = \frac{\pi}{2}$ , $\cos\delta = 0$ : $A = \sqrt{4.0^2 + 3.0^2} = 5.0\,\mathrm{cm}$ . $\tan\phi = \frac{3.0 \times 1}{4.0 + 0} = 0.75$ , so $\phi = 36.9^\circ = 0.64\,\mathrm{rad}$ . The resultant is $x = 5.0\sin(\omega t + 0.64)\,\mathrm{cm}$ .

Worked example 2: the general case

Find the resultant of $x_1 = 3.0\sin\omega t$ and $x_2 = 2.0\sin\left(\omega t + \frac{\pi}{3}\right)$ , in centimetres. (These are the motions drawn above.)

Solution

$A = \sqrt{3.0^2 + 2.0^2 + 2 \times 3.0 \times 2.0 \times \cos 60^\circ} = \sqrt{9.0 + 4.0 + 6.0} = \sqrt{19} = 4.36\,\mathrm{cm}$ .

$\tan\phi = \frac{2.0\sin 60^\circ}{3.0 + 2.0\cos 60^\circ} = \frac{1.732}{4.0} = 0.433$ , so $\phi = 23.4^\circ = 0.41\,\mathrm{rad}$ .

The resultant, $x = 4.36\sin(\omega t + 0.41)\,\mathrm{cm}$ , lies between the two in phase, closer to the larger motion, as the phasor diagram shows.

Worked example 3: how good is the cancellation?

A headphone produces anti-noise of exactly the same amplitude as the noise, but the circuit's delay makes the phase difference $170^\circ$ instead of $180^\circ$ . What fraction of the noise amplitude is left?

Solution

With $A_{1} = A_{2} = A_{0}$ : $A^2 = 2A_0^2(1 + \cos\delta) = 4A_0^2\cos^2\frac{\delta}{2}$ , so $A = 2A_0\left|\cos\frac{\delta}{2}\right| = 2A_0\cos 85^\circ = 0.174A_0$ . A phase error of only $10^\circ$ leaves 17% of the amplitude. By comparison, a perfect phase but an amplitude 5% too small leaves $∣ A_{1} - A_{2} ∣ = 0.05 A_{0}$ , only 5%. Accurate timing matters more than accurate loudness, and the timing is hardest at high frequencies, where $10^\circ$ is a very short delay. That is why noise cancelling works best on low, steady sounds such as engine hum.

Activity 1.3: adding phasors by scale drawing

You need: squared paper, a ruler, a protractor and a sharp pencil.

Choose a scale, for example 1 cm on paper for 1 cm of amplitude. Draw $A_{1} = 5.0$ along a horizontal reference line.
From the tip of $A_{1}$ , draw $A_{2} = 3.0$ at $60^\circ$ to the reference.
Join the start of $A_{1}$ to the tip of $A_{2}$ . Measure this resultant and its angle to the reference.
Check your drawing with the formula; you should find $A = 7.0$ and $\phi \approx 22^\circ$ .
Repeat with $\delta = 0^\circ, 90^\circ, 120^\circ$ and $180^\circ$ , and plot $A$ against $\delta$ . Describe the shape of your graph.

Warning

Common mistakes. (1) Adding amplitudes directly, $A = A_{1} + A_{2}$ , whatever the phase: this is true only when $\delta = 0$ . (2) Using $-2A_1A_2\cos\delta$ , as in the cosine rule for the angle inside the triangle; with $\delta$ defined as the phase difference the sign is $+$ . (3) Taking $\tan\phi = \frac{A_2}{A_1}$ for every case; that holds only in quadrature. (4) Expecting the sum of two different frequencies to be SHM.

Tip

Draw the phasors before you calculate. The picture tells you at once whether the answer should be bigger or smaller than the larger amplitude, and which motion it leans towards, so you can catch a sign error in the formula.

How Do Noise Canceling Headphones Work? (Branch Education). Watch for the anti-noise signal that is half a cycle out of step with the noise, and for why low, steady sounds cancel best.

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Practice

State the principle of superposition for two SHMs along the same line.
Find the resultant amplitude and phase of $x_1 = 6.0\sin\omega t$ and $x_2 = 8.0\sin\left(\omega t + \frac{\pi}{2}\right)$ .
Two SHMs of amplitude 4.0 cm each differ in phase by $\frac{2\pi}{3}$ . Find the resultant amplitude.
Find the resultant of $x_1 = 5.0\sin\omega t$ and $x_2 = 3.0\sin\left(\omega t + \frac{\pi}{3}\right)$ .
Write $x = 3\sin\omega t + 4\cos\omega t$ in the form $A\sin(\omega t + \phi)$ .
Use the formula for $A$ to show that $|A_1 - A_2| \le A \le A_1 + A_2$ .
Anti-noise has the same amplitude as the noise and a phase difference of $175^\circ$ . What fraction of the amplitude remains?
Gatete says: "An SHM of amplitude 3 cm plus one of 4 cm always gives 7 cm." Evaluate his statement, and give the range of possible results.

Show answersHide answers

The resultant displacement at any instant is the sum of the displacements that each SHM would produce alone.
$A = \sqrt{36 + 64} = 10\,\mathrm{cm}$ ; $\tan\phi = \frac{8.0}{6.0}$ , $\phi = 53.1^\circ$ (0.93 rad).
$A^2 = 16 + 16 + 32\cos 120^\circ = 16$ , so $A = 4.0\,\mathrm{cm}$ .
$A = \sqrt{25 + 9 + 30 \times 0.5} = 7.0$ ; $\tan\phi = \frac{3.0 \times 0.866}{5.0 + 1.5} = 0.400$ , $\phi = 21.8^\circ$ (0.38 rad).
$4\cos\omega t = 4\sin\left(\omega t + \frac{\pi}{2}\right)$ , so $A = 5$ , $\tan\phi = \frac{4}{3}$ : $x = 5\sin(\omega t + 0.93)$ .
$\cos\delta$ lies between $- 1$ and $+ 1$ , so $A^{2}$ lies between $(A_{1} - A_{2})^{2}$ and $(A_{1} + A_{2})^{2}$ .
$2\cos 87.5^\circ = 0.087$ : about 9% of the amplitude remains.
Wrong in general: 7 cm happens only when the two are in phase. The result can be anything from $4 - 3 = 1\,\mathrm{cm}$ (antiphase) to 7 cm, for example 5 cm in quadrature.

Key points

Same-frequency SHMs along one line add to an SHM of the same frequency.
$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\delta}$ , $\tan\phi = \frac{A_2\sin\delta}{A_1 + A_2\cos\delta}$ ; phasors give the same result by vector addition.
In phase: $A_{1} + A_{2}$ . Antiphase: $∣ A_{1} - A_{2} ∣$ . Quadrature: $\sqrt{A_1^2 + A_2^2}$ .

1Silence made from sound

2The principle of superposition

Deriving the resultant

3Adding with phasors

Three special cases

4Activity 1.3: adding phasors by scale drawing

Silence made from sound

The principle of superposition

Adding with phasors

Activity 1.3: adding phasors by scale drawing