Mathematics Advanced • Year 11 • Module 2 • Lesson 12
Phase Shifts and Horizontal Translations
Apply phase shifts to real-world periodic problems — sound interference, noise-cancelling headphones, AC electricity, gear teeth, and pendulum head-starts.
Problem 1 — Two tuning forks slightly out of phase (sound)
Two identical tuning forks both produce a pure tone at $440\,\text{Hz}$ (concert A). The first fork's sound pressure is modelled by $p_1(t) = \sin(880\pi t)$. The second fork is struck a fraction of a second later and its pressure is $p_2(t) = \sin(880\pi(t - 0.001))$ where $t$ is in seconds.
Set up: What are we solving for?
(i) State the amplitude, period (in seconds), and phase shift (in seconds) of $p_2$ relative to $p_1$. 2 marks
(ii) Convert the $0.001$ s phase shift into a phase shift expressed as a fraction of one period. 2 marks
(iii) Two pure tones are said to be "in phase" if their phase shift is a whole number of periods, "anti-phase" if it is half a period. Is $p_2$ in phase, anti-phase, or neither with $p_1$? Justify with a single calculation. 2 marks
Stuck? Period $T = 2\pi / b$. Express the time delay as a multiple of $T$.Problem 2 — Noise-cancelling headphones (anti-phase)
An aircraft engine produces a steady hum modelled by $N(t) = 2\sin(100\pi t)$ (units: pressure amplitude). Noise-cancelling headphones detect this and produce a counter-wave $C(t)$ such that $N(t) + C(t) = 0$ for all $t$.
Set up: What are we solving for?
(i) Find $C(t)$ in the form $a\sin(b(t - c))$. (There are two natural answers — use the smallest positive $c$.) 3 marks
(ii) What is the phase shift of $C$ relative to $N$, expressed as a fraction of one period? 2 marks
(iii) Engineers also use the equivalent description: $C(t) = -N(t)$. Show that $-N(t) = 2\sin(100\pi(t - c))$ for some value of $c$ you computed in (i). 2 marks
Problem 3 — Comparing two AC supplies (electricity)
Australian mains AC voltage is modelled as $V_{\text{AU}}(t) = 325\sin(100\pi t)$ (peak amplitude $325$ V, frequency $50$ Hz, $t$ in seconds). A second device runs on a phase-shifted backup line: $V_{\text{BU}}(t) = 325\sin(100\pi t - \pi/3)$.
Set up: What are we solving for?
(i) Factor $V_{\text{BU}}$ into the form $325\sin(100\pi(t - c))$ and find $c$ (in seconds, exact). 3 marks
(ii) Convert $c$ to milliseconds (3 dp). 1 mark
(iii) If $V_{\text{AU}}$ reaches its first positive peak at $t = 1/200\,\text{s}$ (i.e. quarter-period), at what time does $V_{\text{BU}}$ reach its first positive peak? Show your reasoning. 3 marks
Stuck? A right shift by $c$ delays every event of the original graph by $c$.Problem 4 — Gear teeth heights as a function of rotation
The height of a tooth on a meshing gear (above the gear's axis) is well-approximated by $h(\theta) = 5 + 0.4\cos(20(\theta - 0.05))$ centimetres, where $\theta$ is the angle of rotation in radians.
Set up: What are we solving for?
(i) State the midline, amplitude, period, and phase shift. Then express the period in degrees of rotation. 3 marks
(ii) The period corresponds to one tooth's spacing. How many teeth has the gear? (Use $360^\circ$ ÷ period-in-degrees, rounded to nearest integer.) 2 marks
(iii) The mating gear has equation $h_2(\theta) = 5 + 0.4\cos(20(\theta - 0.05 - \pi/20))$. By how much (in radians) does $h_2$ lag $h_1$? Why does this lag matter mechanically? 2 marks
Problem 5 — Two pendulums with a head-start
Two identical pendulums of length $1\,\text{m}$ swing in the same plane. Pendulum A starts from rest at its central position at $t = 0$ and its displacement is $x_A(t) = 0.2\sin(\pi t)$ metres. Pendulum B is identical but was released $0.5\,\text{s}$ before $t = 0$.
Set up: What are we solving for?
(i) Write $x_B(t)$ in the form $a\sin(b(t - c)) + d$. 2 marks
(ii) Find $x_B(0)$ — where is pendulum B at $t = 0$? 2 marks
(iii) A student claims pendulum B is "in phase" with A because they have the same amplitude and period. Evaluate this claim — what extra condition would have to hold for two waves with the same amplitude and period to be in phase, and does pendulum B meet it? 2 marks
Stuck on (iii)? "In phase" means the phase shift (modulo period) is zero. Period here is $2$ s.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Tuning forks
Set up. Read parameters from $p_2$, convert the time shift to a fraction of the period, classify as in-phase / anti-phase / neither.
(i) $p_2(t) = \sin(880\pi(t - 0.001))$. Amplitude $1$; period $T = 2\pi/(880\pi) = $ $1/440$ s $\approx 2.27\,\text{ms}$; phase shift $0.001$ s right (a delay).
(ii) Phase shift as fraction of period $= 0.001 / (1/440) = 0.001 \times 440 = $ $0.44$ periods.
(iii) $0.44$ is neither a whole number nor exactly $0.5$, so $p_2$ is neither in-phase nor anti-phase with $p_1$. (It's close to anti-phase, but not exactly.)
Problem 2 — Noise-cancelling headphones
Set up. $C(t) = -N(t) = -2\sin(100\pi t)$. Rewrite this as a phase-shifted sine of the same amplitude.
(i) Using $-\sin\theta = \sin(\theta - \pi)$ (or equivalently $\sin(\theta + \pi)$), $C(t) = 2\sin(100\pi t - \pi) = 2\sin(100\pi(t - 1/100))$. Smallest positive $c$: $c = 1/100$ s $= 0.01$ s. So $C(t) = $ $2\sin(100\pi(t - 0.01))$.
(ii) Period $T = 2\pi/(100\pi) = 0.02$ s. Phase shift $/ $ period $= 0.01 / 0.02 = $ $1/2$ of a period (i.e. anti-phase — the textbook definition of perfect cancellation).
(iii) $-N(t) = -2\sin(100\pi t) = 2\sin(100\pi t - \pi) = 2\sin(100\pi(t - 0.01))$. So $-N(t)$ matches the form with $c = 0.01$. ✓
Problem 3 — AC supplies
Set up. Convert the unfactored phase into factored form, then in seconds, then find the new peak time.
(i) $100\pi t - \pi/3 = 100\pi(t - (\pi/3)/(100\pi)) = 100\pi(t - 1/300)$. So $c = 1/300\,\text{s}$.
(ii) $c = 1/300\,\text{s} = $ $3.333$ ms (to 3 dp).
(iii) $V_{\text{AU}}$ peaks at $t = 1/200$ s. $V_{\text{BU}}$ peaks $c = 1/300$ s later: $1/200 + 1/300 = 3/600 + 2/600 = $ $5/600 = 1/120$ s $\approx 8.33$ ms.
Problem 4 — Gear teeth
Set up. Read the model's parameters, convert period to degrees, count teeth, then compare two gears' phase shifts.
(i) Midline $5$ cm, amplitude $0.4$ cm, period $= 2\pi / 20 = \pi/10$ rad. Phase shift $0.05$ rad right. In degrees: period $= (\pi/10) \times (180/\pi) = $ $18^\circ$.
(ii) Teeth $= 360^\circ / 18^\circ = $ $20$ teeth.
(iii) Additional shift in $h_2$ relative to $h_1$ is $\pi/20$ rad (i.e. $9^\circ$, exactly half a tooth spacing). Mechanically this is the meshing offset — tooth peaks of one gear align with tooth valleys of the other, which is exactly what enables them to mesh smoothly.
Problem 5 — Racing pendulums
Set up. B was released $0.5$ s before A, so its "$t = 0$ event" is at $t = -0.5$ in the global clock — i.e. $x_B(t) = x_A(t + 0.5) = 0.2\sin(\pi(t + 0.5))$.
(i) $x_B(t) = 0.2\sin(\pi(t - (-0.5))) = $ $0.2\sin(\pi(t + 0.5))$ (or equivalently $0.2\sin(\pi t + \pi/2)$). Amplitude $0.2$, period $2$ s, phase shift $0.5$ s left, vertical shift $0$.
(ii) $x_B(0) = 0.2\sin(\pi \cdot 0.5) = 0.2 \sin(\pi/2) = 0.2 \times 1 = $ $0.2$ m (i.e. B is at its maximum displacement at $t = 0$, since it had a quarter-period head start).
(iii) The claim is incorrect. "In phase" requires not just matching amplitude and period but also a phase shift that is an integer multiple of the period. Period is $2$ s; B has phase shift $-0.5$ s, which is $-1/4$ of a period — neither $0$ nor any integer multiple. So B is not in phase with A (in fact it's at quadrature — $90^\circ$ phase difference).