Mathematics Advanced • Year 11 • Module 2 • Lesson 15
Review and Connections
Apply the whole module to one integrated context per problem — water wheel, music tuning, biorhythm models, harbour scheduling, and a multi-step pendulum with identity check.
Problem 1 — Water-wheel bucket (model + solve)
A water wheel rotates so the height of one bucket above the water surface is $h(t) = 2\sin\!\left(\dfrac{\pi t}{4}\right) + 1$ metres, where $t$ is in seconds.
Set up: What are we solving for?
(i) Find the maximum and minimum heights, and the period. 2 marks
(ii) Find the first time $t > 0$ when the bucket is at $h = 2$ m. 2 marks
(iii) The bucket is "submerged" when $h \leq 0$. Find the interval of $t \in [0, 8]$ during which it is submerged (3 dp). State one limitation of the model. 3 marks
Stuck on (iii)? Solve $2\sin(\pi t/4) + 1 = 0$.Problem 2 — Two musical notes (phase + frequency)
Two notes are played at the same time. Note A is concert A: $p_A(t) = \sin(880\pi t)$. Note B is concert A an octave higher: $p_B(t) = \sin(1760\pi(t - 0.0001))$ (note: doubled frequency, slight delay).
Set up: What are we solving for?
(i) State the frequency (Hz) and period (in seconds) of each note. 2 marks
(ii) Express the $0.0001$ s delay of $p_B$ as a fraction of $p_B$'s period. 2 marks
(iii) A piano tuner uses the identity $\sin(\theta) + \sin(\phi) = 2\sin\!\left(\dfrac{\theta+\phi}{2}\right)\cos\!\left(\dfrac{\theta-\phi}{2}\right)$ (sum-to-product) to spot beat frequencies. Without evaluating it numerically, describe in one sentence what kind of pattern $p_A + p_B$ would produce graphically and why an octave-pair "sounds good" — i.e. give the qualitative reason from the periodicity of the components. 2 marks
Problem 3 — Biorhythm "physical cycle" (pseudoscience, fun maths)
The biorhythm idea (pseudoscientific but mathematically interesting) models a person's "physical state" as $P(t) = \sin\!\left(\dfrac{2\pi t}{23}\right)$, where $t$ is days since birth and the cycle period is $23$ days. (Values in $[-1, 1]$; positive is "high", negative is "low".)
Set up: What are we solving for?
(i) A person is $7300$ days old. Find $P(7300)$ to 3 dp. Hint: compute $7300 \bmod 23$ first to reduce the angle. 3 marks
(ii) Find all $t \in [0, 23]$ at which $P(t) = 0$ (the "critical" days). 2 marks
(iii) Identify two reasons why a single sine wave is a poor model for actual human physical state. 2 marks
Problem 4 — Harbour ship-arrival window
A harbour's tide is $h(t) = 4 + 2\cos\!\left(\dfrac{2\pi t}{12}\right)$ metres, $t$ in hours since 6 am (high tide at 6 am). A bulk carrier needs $h \geq 5\,\text{m}$ to dock safely.
Set up: What are we solving for?
(i) State the period and the high and low tide values. 2 marks
(ii) Find the times $t > 0$ in $[0, 12]$ at which $h(t) = 5$. Give in exact form. 3 marks
(iii) The harbour's day-shift runs 6 am to 6 pm. By symmetry of cosine about $t = 0$, give the total docking window in this 12 h period and express it as a clock time range. 3 marks
Stuck on (iii)? The set $\{t : h(t) \geq 5\}$ is symmetric about $t = 0$, but the day-shift is $t \in [0, 12]$ — combine the post-6am window with the pre-6pm window.Problem 5 — Pendulum motion: model and identity check
A pendulum's horizontal displacement is $x(t) = 0.1\sin(2t) + 0.05\cos(2t)$ metres.
Set up: What are we solving for?
(i) Show that $x(t)$ can be written as $R\sin(2t + \phi)$ where $R = \sqrt{0.1^2 + 0.05^2}$ and $\tan\phi = 0.05/0.1$. Give $R$ in exact form and $\phi$ to 3 dp. 3 marks
(ii) Using your form, state the amplitude (in metres) and period (in seconds) of $x(t)$. 2 marks
(iii) Verify your form by computing $x(0)$ from both expressions and showing they agree. 2 marks
Stuck on (i)? Use the auxiliary-angle identity: $a\sin\theta + b\cos\theta = R\sin(\theta + \phi)$ with $R = \sqrt{a^2 + b^2}$ and $\tan\phi = b/a$.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Water wheel
Set up. Parameter extraction, then solve for one threshold and one inequality.
(i) Max $= 2 + 1 = $ $3$ m; min $= -2 + 1 = $ $-1$ m; period $= 2\pi/(\pi/4) = $ $8$ s.
(ii) $2\sin(\pi t/4) + 1 = 2 \Rightarrow \sin(\pi t/4) = 1/2 \Rightarrow \pi t/4 = \pi/6 \Rightarrow t = $ $2/3$ s.
(iii) $h \leq 0 \Rightarrow \sin(\pi t/4) \leq -1/2$. In $[0, 8]$ (one period), this holds for $\pi t/4 \in [\pi + \pi/6, 2\pi - \pi/6] = [7\pi/6, 11\pi/6]$, so $t \in [(4/\pi)(7\pi/6), (4/\pi)(11\pi/6)] = [14/3, 22/3] \approx $ $[4.667, 7.333]$ s. Limitation: the bucket cannot physically reach $-1$ m below water on a real wheel (the model exceeds the physical range; in practice the bucket scoops at the bottom and lifts a fixed-volume amount of water).
Problem 2 — Two notes
Set up. Extract frequency/period, normalise the time delay, then describe the sum qualitatively.
(i) A: $b = 880\pi \Rightarrow$ freq $= b/(2\pi) = $ $440$ Hz, period $= 1/440$ s. B: $b = 1760\pi \Rightarrow$ freq $= $ $880$ Hz (one octave above), period $= 1/880$ s.
(ii) Delay $0.0001$ s as fraction of B's period: $0.0001 / (1/880) = 0.088 \approx $ $8.8\%$ of one period.
(iii) $p_A + p_B$ produces a complex but periodic wave with period $1/440$ s (the longer of the two periods, since $880$ Hz is exactly twice $440$ Hz — the higher tone completes two cycles in one period of the lower tone). The two frequencies share a common fundamental period, so the combined wave is exactly periodic — that integer ratio is precisely why octaves sound consonant.
Problem 3 — Biorhythm
Set up. Reduce a huge time argument modulo the period, then solve sine = 0 in one period.
(i) $7300 \div 23 = 317$ remainder $9$ (since $317 \times 23 = 7291$ and $7300 - 7291 = 9$). So $P(7300) = \sin(2\pi \times 9/23) = \sin(18\pi/23) \approx \sin(2.4593) \approx $ $0.631$ (Q2 — sine positive, falling).
(ii) $\sin(2\pi t/23) = 0 \Rightarrow 2\pi t/23 = n\pi \Rightarrow t = 23n/2$. In $[0, 23]$: $t = $ $0, 11.5, 23$ days.
(iii) Two of (any reasonable): (a) human physical state depends on many factors (sleep, diet, illness) — not just elapsed days. (b) Any "physical cycle" of $23$ days has no biological basis. (c) Different people would have different cycles, not a universal $23$-day period. (d) A sine wave is symmetric; real biological cycles are typically asymmetric (e.g. faster recovery than decline).
Problem 4 — Harbour
Set up. Solve $h = 5$ symmetrically about $t = 0$, then combine the pre-noon and post-noon windows.
(i) Period $= 2\pi/(2\pi/12) = $ $12$ h. High $= 4 + 2 = 6$ m at $t = 0$ (6 am). Low $= 4 - 2 = 2$ m at $t = 6$ (12 pm — noon).
(ii) $4 + 2\cos(2\pi t/12) = 5 \Rightarrow \cos(\pi t/6) = 1/2 \Rightarrow \pi t/6 = \pi/3$ or $5\pi/3$ (within $[0, 2\pi]$) $\Rightarrow t = 2$ or $t = 10$. $t = 2$ h and $t = 10$ h, i.e. 8 am and 4 pm.
(iii) $h(t) \geq 5$ when $\cos(\pi t/6) \geq 1/2$, which holds for $\pi t/6 \in [-\pi/3, \pi/3]$ within one period. By symmetry about $t = 0$, the window in $[0, 12]$ is $[0, 2] \cup [10, 12]$, total length $4$ h. Clock times: 6 am to 8 am, and 4 pm to 6 pm.
Problem 5 — Pendulum + auxiliary angle
Set up. Apply the auxiliary angle identity to combine the two terms into a single sinusoid, then read parameters and verify.
(i) With $a = 0.1$ (sine coeff) and $b = 0.05$ (cosine coeff): $R = \sqrt{0.01 + 0.0025} = \sqrt{0.0125} = \sqrt{1/80} = 1/(4\sqrt{5}) = $ $\sqrt{5}/20 \approx 0.1118$ m. $\tan\phi = 0.05/0.1 = 0.5 \Rightarrow \phi = \arctan(0.5) \approx $ $0.4636$ rad. So $x(t) = $ $(\sqrt{5}/20)\sin(2t + 0.4636)$ (approximately).
(ii) Amplitude $= R = \sqrt{5}/20 \approx $ $0.1118$ m. Period $= 2\pi/2 = $ $\pi$ s $\approx 3.142$ s.
(iii) Original: $x(0) = 0.1\sin(0) + 0.05\cos(0) = 0 + 0.05 = $ $0.05$ m. Auxiliary form: $x(0) = (\sqrt{5}/20)\sin(0.4636) \approx 0.1118 \times 0.4472 \approx $ $0.0500$ m. ✓ They agree.