Modelling with Functions — Periodic Phenomena and Logarithmic Scales
From ocean tides to earthquake magnitudes, the same mathematical structures describe phenomena that span vastly different scales. This lesson builds trig models for periodic real-world events and introduces the logarithmic scales that make enormous numerical ranges manageable.
A factory measures 95 dB; the safe limit is 85 dB. The factory says it's only 10 units over — about 10% more. Before any formulas, write your gut response to these:
- Is "10 units more" the same as "10% more intense" in sound? Why or why not?
- What do you think the decibel scale is actually measuring?
- If pH 3 and pH 7 both seem like small numbers — are they similar levels of acidity?
Two big ideas power this lesson: trig functions model repeating phenomena, and logarithms compress enormous ranges into manageable numbers. Both are about making the invisible structure of real data visible.
Periodic model: $f(t) = a\sin(bt + c) + d$ — amplitude $a$, angular frequency $b = 2\pi/T$, phase shift $c$, vertical shift $d$. Log scale: each unit step equals a multiplicative factor — on the decibel scale, +10 dB means ×10 in intensity.
Key facts
- Parameters $a$, $b$, $c$, $d$ in $f(t) = a\sin(bt+c)+d$ and how each relates to physical features
- Log scale: equal spacing = equal ratio; each step on the dB/Richter/pH scale is multiplicative
- Formulas: $L = 10\log_{10}(I/I_0)$, $M = \log_{10}(A/A_0)$, pH $= -\log_{10}[\text{H}^+]$, Pogson's magnitude relation
Concepts
- Why log scales are used when data spans many orders of magnitude
- Why the factory's "10 units = 10%" argument is wrong (dB is logarithmic, not linear)
- When graphical approaches reveal multiple solutions that algebra alone might miss
Skills
- Build a trig model from periodic data (period, amplitude, phase, midline)
- Solve trig equations algebraically using inverse trig within a given domain
- Apply dB, Richter, stellar magnitude and pH formulas to find unknowns
- Compare values on log scales and interpret the multiplicative difference
Any quantity that rises and falls in a regular repeating pattern can be modelled by a transformed sine or cosine. The four parameters $a$, $b$, $c$, $d$ map directly onto observable features of the context.
The general model: $h(t) = a\sin(bt + c) + d$
- Amplitude $a$: $a = \dfrac{\text{max} - \text{min}}{2}$ — half the total range of the phenomenon.
- Vertical shift (midline) $d$: $d = \dfrac{\text{max} + \text{min}}{2}$ — the average value around which it oscillates.
- Period $T$ → angular frequency $b$: $b = \dfrac{2\pi}{T}$. If the cycle repeats every $T$ units of time, $b$ is determined by this formula.
- Phase shift $c$: Adjust so the model matches where the cycle starts. If the maximum occurs at $t = 0$, use cosine (or set $c = \pi/2$ in the sine form).
Midline: $d = (8 + 2)/2 = 5$. Amplitude: $a = (8 - 2)/2 = 3$.
Period: $T = 12$ hr (from high to low is half a cycle, so full period = 12 hr). $b = 2\pi/12 = \pi/6$.
Since max occurs at $t = 0$: use cosine directly — $h(t) = 3\cos\!\left(\dfrac{\pi t}{6}\right) + 5$.
(Equivalently, $h(t) = 3\sin\!\left(\dfrac{\pi t}{6} + \dfrac{\pi}{2}\right) + 5$.)
$h(t) = a\sin(bt+c)+d$: $a=(\text{max}-\text{min})/2$, $d=(\text{max}+\text{min})/2$, $b=2\pi/T$.; Read the period $T$ directly from context (time for one complete cycle).
Pause — copy the parameter formulas: $a = (\text{max}-\text{min})/2$, $d = (\text{max}+\text{min})/2$, $b = 2\pi/T$ — used to build $h(t) = a\sin(bt+c)+d$ from a described periodic context — into your book.
Quick check: A Ferris wheel of radius 10 m has its centre 12 m above the ground and completes one revolution every 60 seconds. Which model correctly gives the height $h$ (in metres) of a rider starting at the bottom?
We just saw how to build the model $h(t) = a\sin(bt+c)+d$ from maximum, minimum and period. That raises a question: once you have the model, solving "when does $h(t) = k$?" can give zero, one, two, or more solutions — how do you count and find them all without missing any? This card answers it → sketch the graph and draw the horizontal line $y = k$ to count intersections first, then use algebra (isolate $\sin\theta = k$, find principal value, symmetry, back-substitute) for exact answers.
Trig equations often have multiple solutions within a domain. Sketching the graph first reveals how many intersections exist and prevents missing solutions. Algebra then gives exact values.
Graphical approach:
- Sketch the transformed function over the domain.
- Draw a horizontal line at the target value.
- Count intersections — that is how many solutions exist.
- Read approximate values from the graph; useful for checking algebraic answers.
Algebraic approach:
- Rearrange to isolate the trig ratio: $\sin(\theta) = k$.
- Find the principal value: $\theta_1 = \arcsin(k)$.
- Use symmetry to find all solutions within the domain: $\theta_2 = \pi - \theta_1$ (for sine), or use the unit circle.
- Back-substitute to solve for $t$.
Set up: $500\sin\!\left(\dfrac{\pi t}{12}\right) + 2000 = 2500$
$\Rightarrow \sin\!\left(\dfrac{\pi t}{12}\right) = 1$
$\Rightarrow \dfrac{\pi t}{12} = \dfrac{\pi}{2}$ (principal), so $t = 6$.
In $[0, 24]$: sine reaches 1 only once per period (24 hr) at $t = 6$. Sketch confirms one solution. Answer: $t = 6$ hours.
Sketch trig graph + horizontal line at target value to count solutions before solving.; Algebraic steps: isolate $\sin\theta = k$ → principal value $\arcsin(k)$ → symmetry for second value → back-substitute for $t$.
Pause — copy the two-step approach: sketch the trig function and horizontal line to count solutions; then algebraically isolate $\sin\theta = k$, find the principal value $\arcsin(k)$, use symmetry for the second value, and back-substitute — into your book.
True or false: For the equation $P(t) = 500\sin(\pi t/12) + 2000$, the equation $P = 2500$ has exactly two solutions in the interval $0 \le t \le 24$.
We just saw how to solve trig equations graphically and algebraically. That raises a question: when data spans many orders of magnitude — like pH from $10^{-14}$ to $10^0$, or sound intensity from $10^{-12}$ W/m² to $10^2$ W/m² — how do you display and reason about it without the small values becoming invisible on a linear scale? This card answers it → a logarithmic scale places equal spacing at equal multiplicative ratios, so that each step of 1 on the scale represents a factor of 10 in the original quantity.
When data spans many orders of magnitude — from $10^{-14}$ to $10^0$, for example — a linear scale is useless: everything looks the same. Logarithmic scales solve this by compressing multiplication into addition.
Why we need log scales:
- pH example: pH ranges 0–14, which looks like a narrow interval. But the underlying $[\text{H}^+]$ ranges from $10^0 = 1$ mol/L (pH 0) to $10^{-14}$ mol/L (pH 14) — 14 orders of magnitude, a factor of $10^{14}$. A linear scale would make pH 3–7 indistinguishable.
- Key property: On a log scale, equal spacing = equal ratio. Moving 1 unit on the dB scale = multiplying intensity by 10. Moving 3 units = multiplying by $10^3 = 1000$.
- When to use a log scale: When data spans more than about 2–3 orders of magnitude; when percentage or ratio changes matter more than absolute differences; when exponential growth/decay is being studied.
Log scale: equal spacing = equal ratio (multiply by same factor). Each step of 1 on a $\log_{10}$ scale = ×10 change in original value.; Use when data spans many orders of magnitude or when ratios matter more than differences.
Pause — copy the log-scale principle: equal spacing = equal ratio; each step of 1 on a $\log_{10}$ scale = $\times 10$ change in the original value — into your book.
Fill the blanks: pH $= -\log_{10}[\text{H}^+]$. If $[\text{H}^+] = 10^{-5}$ mol/L, then pH $= -\log_{10}(10^{-5}) = $ . This solution is (pH < 7 means acidic). A solution with pH 3 has a $[\text{H}^+]$ ratio compared to pH 5 of $10^{2} = $ times more acidic.
We just saw that logarithmic scales compress huge ranges by mapping equal ratios to equal spacing. That raises a question: for sound and earthquakes specifically, what are the exact formulas — and what does a "+10 dB" or "+1 magnitude" step actually mean in physical terms? This card answers it → decibels: $L = 10\log_{10}(I/I_0)$ so $+10$ dB multiplies intensity by 10; Richter: $M = \log_{10}(A/A_0)$ so $+1$ magnitude multiplies amplitude by 10.
The decibel and Richter scales both use $\log_{10}$ to compress intensity or amplitude data spanning many orders of magnitude. A small numerical difference on these scales corresponds to an enormous physical difference.
Decibel scale (sound intensity):
$$L = 10\log_{10}\!\left(\frac{I}{I_0}\right) \quad \text{where } I_0 = 10^{-12} \text{ W/m}^2$$
- $I_0 = 10^{-12}$ W/m² is the threshold of human hearing (reference level).
- +10 dB = intensity ×10 (not ×10 in perceived loudness — perception is also logarithmic, so +10 dB sounds roughly twice as loud to humans).
- Worked example: Concert hall at 85 dB. Find $I$: $85 = 10\log_{10}(I/10^{-12})$, so $\log_{10}(I/10^{-12}) = 8.5$, giving $I = 10^{8.5} \times 10^{-12} = 10^{-3.5} \approx 3.16 \times 10^{-4}$ W/m².
- Normal conversation at 60 dB has $I = 10^{-6}$ W/m². Ratio: $10^{-3.5}/10^{-6} = 10^{2.5} \approx 316$ — the concert is about 316 times more intense.
Richter scale (seismic amplitude):
$$M = \log_{10}\!\left(\frac{A}{A_0}\right)$$
- Each whole-number increase in magnitude = ×10 in ground-motion amplitude.
- M 6.0 vs M 8.0: difference of 2 units → ratio of $10^2 = 100$ in ground-motion amplitude.
Decibel: $L = 10\log_{10}(I/I_0)$, $I_0 = 10^{-12}$ W/m². +10 dB = ×10 intensity.; Richter: $M = \log_{10}(A/A_0)$. +1 magnitude = ×10 ground-motion amplitude.
Pause — copy the decibel formula $L = 10\log_{10}(I/I_0)$ where $I_0 = 10^{-12}$ W/m² (+10 dB = ×10 intensity) and the Richter formula $M = \log_{10}(A/A_0)$ (+1 magnitude = ×10 amplitude) into your book.
Quick check: A sound has intensity $I = 10^{-6}$ W/m². Using $L = 10\log_{10}(I/I_0)$ where $I_0 = 10^{-12}$ W/m², its loudness in decibels is:
We just saw that the decibel and Richter scales use $10\log_{10}$ and $\log_{10}$ to compress physical intensity into small manageable numbers. That raises a question: the same logarithmic reasoning applies in astronomy and chemistry — what are the formulas for stellar magnitude and pH, and what physical ratio corresponds to a one-unit change on each scale? This card answers it → stellar: $m_1 - m_2 = -2.5\log_{10}(F_1/F_2)$ (5 magnitudes = ×100 brightness); pH: $-\log_{10}[\text{H}^+]$ (1 unit = ×10 in $[\text{H}^+]$).
The same logarithmic reasoning extends to astronomy (stellar brightness) and chemistry (acidity). In each case: set up the log equation, apply log laws, solve for the unknown.
Apparent magnitude (stellar brightness):
$$m_1 - m_2 = -2.5\log_{10}\!\left(\frac{F_1}{F_2}\right)$$
- Lower magnitude number = brighter star (the scale is inverted).
- A difference of 5 magnitudes = flux ratio of $10^{5/2.5} = 10^2 = 100$ (Pogson's ratio).
- Example: Star A has $m = 1$, Star B has $m = 6$. Flux ratio: $1 - 6 = -2.5\log_{10}(F_A/F_B)$, so $F_A/F_B = 10^{5/2.5} = 100$. Star A is 100× brighter.
pH scale:
$$\text{pH} = -\log_{10}[\text{H}^+]$$
- Higher pH = lower $[\text{H}^+]$ = less acidic (more basic).
- pH 3 vs pH 5: $[\text{H}^+]$ ratio = $10^{5-3} = 10^2 = 100$. pH 3 is 100× more acidic.
- Each unit of pH = ×10 change in $[\text{H}^+]$.
Stellar magnitude: $m_1 - m_2 = -2.5\log_{10}(F_1/F_2)$. Lower $m$ = brighter. Difference of 5 magnitudes = ×100 brightness.; pH $= -\log_{10}[\text{H}^+]$. Each pH unit = ×10 change in $[\text{H}^+]$. pH < 7 acidic, pH > 7 basic.
Pause — copy the stellar magnitude formula $m_1 - m_2 = -2.5\log_{10}(F_1/F_2)$ (lower $m$ = brighter; 5 magnitudes = ×100) and the pH formula $\text{pH} = -\log_{10}[\text{H}^+]$ (1 pH unit = ×10 change in $[\text{H}^+]$) into your book.
Match each log-scale formula with its application area:
Worked examples · 3 problems
The height of water at a harbour is modelled by a periodic function. High tide is 7.5 m at $t = 0$ hours; low tide is 1.5 m at $t = 6$ hours. (a) Write a model of the form $h(t) = a\cos(bt) + d$. (b) Find the height at $t = 4$ hours. (c) Find all times in $0 \le t \le 12$ when $h = 5.25$ m.
$d = (7.5 + 1.5)/2 = 4.5$ (midline)
$a = (7.5 - 1.5)/2 = 3$ (amplitude)
$T = 12$ hr (high→low = 6 hr = half period), so $b = 2\pi/12 = \pi/6$
Max at $t=0$: use cosine → $h(t) = 3\cos\!\left(\dfrac{\pi t}{6}\right) + 4.5$
$h(4) = 3\cos\!\left(\dfrac{\pi \cdot 4}{6}\right) + 4.5 = 3\cos\!\left(\dfrac{2\pi}{3}\right) + 4.5$
$= 3 \times (-0.5) + 4.5 = -1.5 + 4.5 = 3.0$ m
$3\cos(\pi t/6) + 4.5 = 5.25$
$\cos(\pi t/6) = 0.25$
$\pi t/6 = \arccos(0.25) \approx 1.318$ rad
$t \approx 6 \times 1.318/\pi \approx 2.52$ hr (first solution)
Second: $\pi t/6 = 2\pi - 1.318 \approx 4.965$, so $t \approx 9.48$ hr
Both in $[0, 12]$. ✓
A quiet library measures 30 dB. A busy café measures 70 dB. Using $L = 10\log_{10}(I/I_0)$ with $I_0 = 10^{-12}$ W/m²: (a) Find the intensity $I$ for each location. (b) How many times more intense is the café than the library?
Library (30 dB): $30 = 10\log_{10}(I/10^{-12})$
$\log_{10}(I/10^{-12}) = 3$, so $I/10^{-12} = 10^3$
$I_{\text{library}} = 10^3 \times 10^{-12} = 10^{-9}$ W/m²
Café (70 dB): $70 = 10\log_{10}(I/10^{-12})$
$I_{\text{café}} = 10^7 \times 10^{-12} = 10^{-5}$ W/m²
$\dfrac{I_{\text{café}}}{I_{\text{library}}} = \dfrac{10^{-5}}{10^{-9}} = 10^4 = 10\,000$
Alternatively using level difference:
$\Delta L = 70 - 30 = 40$ dB
ratio $= 10^{40/10} = 10^4 = 10\,000$
Earthquake A registers $M = 5.0$. Earthquake B has ground-motion amplitude 50 times greater than earthquake A. (a) Find the magnitude of earthquake B. (b) How does this compare numerically to $M = 5.0$, and what does this suggest about the log scale?
$M_A = \log_{10}(A_A/A_0) = 5.0$
$A_A/A_0 = 10^5$
$A_B = 50 \times A_A$, so $A_B/A_0 = 50 \times 10^5$
$M_B = \log_{10}(50 \times 10^5) = \log_{10}(50) + 5$
$= \log_{10}(50) + 5 \approx 1.699 + 5 = 6.70$
$M_B - M_A = 6.70 - 5.0 = 1.70$ on the Richter scale
Amplitude ratio: $50\times$ appears as only 1.70 units
If it were 100× greater: $M_B = 5 + \log_{10}(100) = 7.0$ (only 2 units up)
If it were 1000× greater: $M_B = 5 + 3 = 8.0$ (only 3 units up)
Multiple choice · 5 questions
Q1. A Ferris wheel has radius 10 m, its centre is 12 m above the ground, and it completes one revolution every 60 seconds. A rider starts at the bottom. The correct height model $h(t)$ in metres is:
Q2. A sound has intensity $I = 10^{-6}$ W/m². Using $L = 10\log_{10}(I/I_0)$ where $I_0 = 10^{-12}$ W/m², its loudness in decibels is:
Q3. pH 4 compared to pH 7: pH 4 is how many times more acidic?
Q4. A seismic reading increases from magnitude 5 to magnitude 7 on the Richter scale. The ground-motion amplitude increases by a factor of:
Q5. A temperature oscillates with max 32°C, min 18°C, period 24 hours, and reaches its maximum at $t = 14$ hours. The amplitude of the model is:
The factory's argument was mathematically wrong. A difference of 10 dB means the intensity ratio is $10^{10/10} = 10$ — the factory is 10 times more intense than the safe limit, not 10% more. The decibel scale is logarithmic: equal-looking steps correspond to multiplicative changes, not additive ones. The same is true for pH, Richter magnitude and stellar magnitude. Now that you've worked through the full lesson, revisit your gut answers:
SA 1. A buoy bobs with height $h(t) = 1.5\sin\!\left(\dfrac{\pi t}{8}\right) + 3$ metres above the seabed, where $t$ is time in seconds.
(a) State the period and amplitude of the motion. (1 mark)
(b) Find the maximum and minimum height of the buoy. (1 mark)
(c) Find the first two times $t > 0$ at which $h = 3.75$ m. (2 marks)
SA 2. A sound engineer records two sounds: one at 40 dB and another at 70 dB. Using $L = 10\log_{10}(I/I_0)$:
(a) Find the intensity of each sound in W/m² (take $I_0 = 10^{-12}$ W/m²). (2 marks)
(b) Find the ratio of their intensities. (1 mark)
(c) The engineer claims "70 dB is almost twice as loud as 40 dB." Evaluate this claim mathematically. (1 mark)
SA 3. An earthquake registers $M = 5.8$ on the Richter scale ($M = \log_{10}(A/A_0)$). A second tremor has ground-motion amplitude 25 times greater.
(a) Show that the magnitude of the second tremor is $5.8 + \log_{10}(25)$. (2 marks)
(b) Find this magnitude correct to 2 decimal places. (1 mark)
(c) Explain why scientists prefer logarithmic scales for reporting seismic data. (2 marks)
📖 Comprehensive answers (click to reveal)
SA 1 (4 marks):
(a) Period $T = 2\pi / (\pi/8) = 16$ s. Amplitude $= 1.5$ m. [1]
(b) Max $= 3 + 1.5 = 4.5$ m. Min $= 3 - 1.5 = 1.5$ m. [1]
(c) $1.5\sin(\pi t/8) + 3 = 3.75 \Rightarrow \sin(\pi t/8) = 0.5$. Principal value: $\pi t/8 = \pi/6 \Rightarrow t = 8/6 = 4/3 \approx 1.33$ s. Second solution in first period: $\pi t/8 = \pi - \pi/6 = 5\pi/6 \Rightarrow t = 40/6 = 20/3 \approx 6.67$ s. [2]
SA 2 (4 marks):
(a) 40 dB: $\log_{10}(I/10^{-12}) = 4 \Rightarrow I = 10^{-8}$ W/m². 70 dB: $\log_{10}(I/10^{-12}) = 7 \Rightarrow I = 10^{-5}$ W/m². [2]
(b) Ratio $= 10^{-5}/10^{-8} = 10^3 = 1000$. [1]
(c) The claim is mathematically inaccurate. 70 dB is not twice 40 dB — it is 1000 times more intense. The decibel scale is logarithmic; you cannot compare dB values by simple multiplication the way you can with linear measurements. [1]
SA 3 (5 marks):
(a) Let $A_1$ be the amplitude of the first tremor. $M_1 = \log_{10}(A_1/A_0) = 5.8$, so $A_1 = A_0 \cdot 10^{5.8}$. $A_2 = 25 A_1 = 25 \cdot A_0 \cdot 10^{5.8}$. $M_2 = \log_{10}(A_2/A_0) = \log_{10}(25 \cdot 10^{5.8}) = \log_{10}(25) + 5.8$. [2]
(b) $M_2 = \log_{10}(25) + 5.8 \approx 1.3979 + 5.8 = 7.20$ (to 2 d.p.). [1]
(c) Earthquake amplitudes span an enormous range — from microseismic tremors ($10^{-8}$ m) to major quakes ($10^{2}$ m or more), roughly 10 orders of magnitude. A linear scale would require numbers in the billions to describe large quakes while microseismic events would be indistinguishable from zero. The log (Richter) scale compresses this range into a small set of manageable numbers, making it easy to compare events and communicate risk to the public. [2]
Five timed questions covering trig models, decibel calculations, Richter comparisons and pH. Gold tier: 90% + speed.
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