Mathematics Advanced • Year 11 • Module 2 • Lesson 10
Graphs of Sine and Cosine
Apply sinusoidal models to real systems: musical tones, daylight hours, tidal heights, Ferris wheels, and sound-wave detection.
Problem 1 — Middle C on a piano (acoustic)
A pure middle-C tone is modelled by the air-pressure variation $P(t) = 0.4 \sin(524 \pi t)$, where $P$ is in pascals and $t$ is in seconds. (The lesson hook notes middle C oscillates 262 times per second.)
Set up: What are we solving for?
(i) Identify the amplitude $a$ and the parameter $b$ in the form $a \sin(bt)$. State the period of $P(t)$ exactly in seconds, then in milliseconds. 3 marks
(ii) Use your period to verify the lesson's claim "middle C oscillates 262 times per second" (i.e. show frequency = 1/period $\approx 262$). 2 marks
(iii) A second piano string plays a tone with twice the frequency (one octave higher). Write the equation $P_2(t)$ assuming the same amplitude. State its new period. 2 marks
Stuck? Revisit lesson § Today's hook — the middle-C example.Problem 2 — Daylight hours in Sydney (astronomical)
The number of daylight hours in Sydney over a year is modelled by $D(t) = 12 + 2.5 \cos\left(\frac{2 \pi t}{365}\right)$, where $D$ is hours and $t$ is days from the summer solstice (Dec 22, taken as $t = 0$).
Set up: What are we solving for?
(i) State the amplitude, period, and range of $D(t)$, then identify which calendar date corresponds to the maximum daylight. 3 marks
(ii) Find $D(0)$ and $D(182.5)$ (the latter is roughly half a year after the solstice, i.e. the winter solstice in June). Comment on what these values represent. 2 marks
(iii) A geographer suggests using $D(t) = 12 + 2.5 \sin\left(\frac{2 \pi t}{365}\right)$ instead. Explain in 1-2 sentences why this changes the location of the maximum but not the amplitude or period. 2 marks
Problem 3 — Sydney harbour tide model (oceanographic)
The tide height in Sydney Harbour for one 12-hour period is modelled by $h(t) = 1.2 + 0.8 \sin\left(\frac{\pi t}{6}\right)$, where $h$ is in metres above a benchmark and $t$ is in hours from low tide.
Set up: What are we solving for?
(i) State the amplitude, period (in hours), and range of $h(t)$. 2 marks
(ii) Find the times $t \in [0, 12]$ (in hours) at which $h(t) = $ midline value $= 1.2$ m. (Hint: solve $\sin\left(\frac{\pi t}{6}\right) = 0$.) 3 marks
(iii) A harbourmaster needs at least 1.5 m clearance for a particular vessel. State whether the high tide $h_{\max}$ exceeds this and by how much. 2 marks
Problem 4 — Ferris wheel height (mechanical)
A Ferris wheel of radius 15 m has its centre 17 m above the ground and completes one revolution every 4 minutes. A passenger's height above the ground after $t$ minutes from boarding (at the bottom) is $H(t) = 17 - 15 \cos\left(\frac{\pi t}{2}\right)$.
Set up: What are we solving for?
(i) State the amplitude, period, midline, and range of $H(t)$. Confirm the model puts the passenger at ground level ($H = 2$ m, the boarding height) at $t = 0$. 3 marks
(ii) Find $H(2)$. Interpret this physically. 2 marks
(iii) Sketch one full revolution of $H(t)$ for $0 \leq t \leq 4$ on labelled axes. Mark the maximum point and the minimum point. 3 marks
Stuck on (i)? Midline = vertical shift = 17 m. Amplitude = 15 m. So range is $[17 - 15, 17 + 15] = [2, 32]$ m.Problem 5 — Reading parameters from a graph (signal processing)
An oscilloscope shows a sinusoidal signal with the following measurements: maximum displacement 5 V, minimum displacement $-1$ V, time for one full cycle 0.02 s. The waveform passes through its midline going upward at $t = 0$.
Set up: What are we solving for?
(i) Find the amplitude $a$, vertical shift $d$, and parameter $b$ (so that the function $a \sin(b t) + d$ has period 0.02 s). 3 marks
(ii) Write the full equation of the waveform in the form $V(t) = a \sin(b t) + d$. 2 marks
(iii) Verify your equation gives $V(0) = d$ (the midline value), confirming the "upward midline crossing at $t = 0$" condition. 2 marks
Stuck on (i)? $d = \frac{\max + \min}{2}$; $a = \frac{\max - \min}{2}$; $b = \frac{2\pi}{\text{period}}$.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Middle C
Set up. We're reading amplitude and $b$ from a sine model, computing period, and extending to a higher-octave waveform.
(i) Amplitude $a = 0.4$ Pa. $b = 524 \pi$. Period $= \frac{2\pi}{b} = \frac{2\pi}{524\pi} = \frac{1}{262}$ s $\approx 3.82$ ms.
(ii) Frequency $= \frac{1}{\text{period}} = 262$ Hz. ✓ Matches the lesson hook's claim.
(iii) Double the frequency means halving the period to $\frac{1}{524}$ s. The new $b = \frac{2\pi}{1/524} = 1048\pi$. So $P_2(t) = \mathbf{0.4 \sin(1048\pi t)}$. New period = $\mathbf{\frac{1}{524}}$ s ($\approx 1.91$ ms).
Problem 2 — Daylight hours
Set up. We're reading parameters of a cosine model and interpreting a phase change.
(i) Amplitude = 2.5 h. Period = $\frac{2\pi}{2\pi/365} = 365$ days (one year). Range = $[12 - 2.5, 12 + 2.5] = \mathbf{[9.5, 14.5]}$ h. Maximum at $t = 0$ (when $\cos(0) = 1$), which is Dec 22 (summer solstice in Sydney). ✓
(ii) $D(0) = 12 + 2.5 \cos(0) = 12 + 2.5 = 14.5$ h (summer solstice, longest day). $D(182.5) = 12 + 2.5 \cos(\pi) = 12 - 2.5 = 9.5$ h (winter solstice, shortest day). Together they confirm the model's swing of 5 hours from solstice to solstice.
(iii) Replacing cos with sin keeps amplitude (2.5) and period (365) unchanged because they depend only on $a$ and $b$. But the maximum shifts: sin reaches 1 at $t = \frac{365}{4} \approx 91$ days (about March 22, autumnal equinox) instead of $t = 0$. The vertical shift, amplitude, and period are all unchanged.
Problem 3 — Sydney tide
Set up. We're reading model parameters, solving a sin = 0 equation, and checking a clearance threshold.
(i) Amplitude = 0.8 m. Period = $\frac{2\pi}{\pi/6} = 12$ h. Range = $[1.2 - 0.8, 1.2 + 0.8] = \mathbf{[0.4, 2.0]}$ m.
(ii) $\sin\left(\frac{\pi t}{6}\right) = 0 \Rightarrow \frac{\pi t}{6} = n\pi \Rightarrow t = 6n$. On $[0, 12]$: $t = \mathbf{0, 6, 12}$ hours. (The model passes the midline three times in a 12-hour window because it starts at midline.)
(iii) High tide $h_{\max} = 1.2 + 0.8 = 2.0$ m. Required clearance 1.5 m. $2.0 - 1.5 = \mathbf{0.5}$ m exceedance — vessel passes with 0.5 m to spare at high tide.
Problem 4 — Ferris wheel
Set up. We're applying $H(t) = d + a \cos(\cdot)$ form, finding height at a specific time, and sketching the cycle.
(i) Amplitude = 15 m. Period = $\frac{2\pi}{\pi/2} = 4$ min. Midline $d$ = 17 m. Range = $[17 - 15, 17 + 15] = \mathbf{[2, 32]}$ m. Check $H(0) = 17 - 15 \cos(0) = 17 - 15 = 2$ m — boarding height. ✓
(ii) $H(2) = 17 - 15 \cos(\pi) = 17 - 15(-1) = 17 + 15 = \mathbf{32}$ m. Interpretation: 2 minutes (half a revolution) after boarding, the passenger is at the top of the wheel — the maximum height.
(iii) Sketch: a cosine-style wave (inverted because of the "$-15 \cos$" term, so it starts at the minimum). Key points: $(0, 2)$ minimum, $(1, 17)$ midline going up, $(2, 32)$ maximum, $(3, 17)$ midline going down, $(4, 2)$ minimum again. Smooth wave between.
Problem 5 — Reading parameters from a graph
Set up. We're extracting $a$, $d$, $b$ from max/min/period data, then writing and verifying the equation.
(i) $d = \frac{5 + (-1)}{2} = \mathbf{2}$ V. $a = \frac{5 - (-1)}{2} = \mathbf{3}$ V. $b = \frac{2\pi}{0.02} = \mathbf{100\pi}$ rad/s.
(ii) $V(t) = \mathbf{3 \sin(100\pi t) + 2}$.
(iii) $V(0) = 3 \sin(0) + 2 = 0 + 2 = \mathbf{2}$ V, which is exactly the midline value $d$. Also, the derivative $V'(0) = 3 \cdot 100\pi \cdot \cos(0) > 0$, so the curve is moving upward at $t = 0$. Both conditions are satisfied. ✓