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).