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.
Today's hook, A factory near a school measures 95 dB. The safe limit is 85 dB. The factory argues it's only 10 units over, barely 10% more. Is that argument mathematically sound? Think about what the decibel scale actually represents before we dig in.
0/5QUESTS
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Orient to mathematical modelling
Test your intuition, anchor the key relationships and set the goals for the lesson.
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Recall, your gut answer first
+5 XP warm-up
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?
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The key relationships you need to own
+5 XP to read
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.
When building a trig model: find the midline $d = (\text{max}+\text{min})/2$ and amplitude $a = (\text{max}-\text{min})/2$ before worrying about $b$ or $c$. These two come directly from the data.
Log scale: ratio, not difference
A difference of 3 on a log scale means a ratio of $10^3 = 1000$, not a difference of 3. This is why pH 4 and pH 7 differ by a factor of 1000 in acidity, even though the numbers look close.
Graphical vs algebraic
For trig problems: sketch the graph first to see how many solutions exist in the domain, then use algebra (inverse trig) to find exact values. The sketch prevents missing solutions.
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What you'll master
Know
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
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
Can do
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
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Key terms
Periodic phenomenonA process that repeats at regular intervals, tides, temperature cycles, sound waves, Ferris wheels. Modelled by trig functions.
Amplitude ($a$)Half the distance between the maximum and minimum values: $a = (\text{max} - \text{min})/2$. Always positive in context.
Period ($T$)The time for one complete cycle. Links to $b$ via $b = 2\pi/T$. Read directly from context (e.g., "one tide every 12 hours").
Phase shift ($c$)Horizontal translation of the trig function. Shifts the starting point of the cycle left or right.
Logarithmic scaleA scale where equal distances represent equal ratios (not equal differences). Each unit step multiplies the original value by a fixed factor (usually 10).
Decibel (dB)A logarithmic unit for sound intensity: $L = 10\log_{10}(I/I_0)$. +10 dB means the intensity is 10 times greater.
Richter scaleLogarithmic scale for earthquake ground-motion amplitude: $M = \log_{10}(A/A_0)$. A magnitude increase of 1 means ×10 in amplitude.
pHNegative log of hydrogen ion concentration: pH $= -\log_{10}[\text{H}^+]$. pH 3 is 10 000× more acidic than pH 7.
Order of magnitudeA factor of 10. Data spanning 6 orders of magnitude ranges across a factor of $10^6 = 1\,000\,000$.
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Build models of periodic phenomena
Translate real measurements into amplitude, period, phase and midline parameters.
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Modelling Periodic Phenomena with Trig
core concept
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).
Ocean tide worked example: High tide 8 m at $t = 0$, low tide 2 m at $t = 6$ hours.
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, $h(t) = 3\sin\!\left(\dfrac{\pi t}{6} + \dfrac{\pi}{2}\right) + 5$.)
Contexts to practise recognising: Tides (high/low, period ~12 hr), temperature variation (summer/winter, period 12 months), Ferris wheel height (top/bottom, period = rotation time), sound wave pressure (crest/trough, period = 1/frequency), rotating turbine angle.
$h(t) = a\sin(bt+c)+d$: $a=(\text{max}-\text{min})/2$, $d=(\text{max}+\text{min})/2$, $b=2\pi/T$.; 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?
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Compare graphical and algebraic approaches
Choose a useful representation and interpret its strengths and limitations.
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Graphical vs Algebraic Approaches
core concept
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 $t$.
Population oscillation example: $P(t) = 500\sin\!\left(\dfrac{\pi t}{12}\right) + 2000$ bacteria, where $t$ is hours. Find when $P = 2500$ in $0 \le t \le 24$.
In $[0, 24]$: sine reaches 1 only once per period (24 hr) at $t = 6$. Sketch confirms one solution. Answer: $t = 6$ hours.
When to use each approach: Use graphical for quick reads of approximate times, counting solutions, and checking reasonableness. Use algebraic for exact answers required in exam working. Always do both on harder problems.
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 $0 \le t \le 24$.
False. $P = 2500$ requires $\sin(\pi t/12) = 1$, which equals 1 only once per period. The period is 24 hours, so there is exactly one solution ($t = 6$) in $[0, 24]$. If the target were less extreme (say $P = 2250$, so $\sin = 0.5$), there would be two solutions per period.
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Interpret logarithmic scales
Explain how logarithms compress multiplicative change into manageable numerical steps.
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Logarithmic Scales, The Big Picture
core concept
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 vs linear scale, key contrast: On a linear scale, 10, 20, 30 are equally spaced. On a $\log_{10}$ scale, 10, 100, 1000 are equally spaced (each is ×10 the previous). The formula $y = \log_{10}(x)$ converts the multiplicative structure of $x$ into the additive structure of $y$, this is why log scales "compress" large ranges.
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.
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Apply logarithmic models
Calculate and interpret decibel, seismic, stellar-magnitude and pH relationships.
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The Decibel and Seismic Scales
core concept
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.
$I_0 = 10^{-12}$ W/m² is the threshold of human hearing (reference level).
+10 dB = intensity ×10. That is all the formula tells you. How much louder a sound seems is a separate, perceptual question that this formula does not answer, so keep your written answers on the intensity ratio.
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.
Back to the factory question: The factory at 95 dB vs safe limit of 85 dB. The difference of 10 dB means the intensity ratio is $10^{(95-85)/10} = 10^1 = 10$. The factory's sound is 10 times more intense than the safe limit, not 10% more. The argument is completely wrong because it treats a logarithmic scale as if it were linear.
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:
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Star Magnitudes and pH
core concept
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.
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 $[\text{H}^+]$.
General approach for all log-scale problems: (1) Write the formula. (2) Substitute known values. (3) Isolate the log expression. (4) Exponentiate both sides (or use log laws). (5) Solve for the unknown. (6) Interpret in context, is the answer a ratio, a concentration, an intensity?
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:
$L = 10\log_{10}(I/I_0)$
$\text{pH} = -\log_{10}[\text{H}^+]$
$M = \log_{10}(A/A_0)$
$m_1 - m_2 = -2.5\log_{10}(F_1/F_2)$
Earthquake ground-motion amplitude
Apparent brightness of stars
Sound intensity in decibels
Acidity of a chemical solution
Worked examples · 3 problems
PROBLEM 1 · BUILDING AND USING A TIDAL MODEL
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.
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Part (a), find parameters $d = (7.5 + 1.5)/2 = 4.5$ (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 → $h(t) = 3\cos\!\left(\dfrac{\pi t}{6}\right) + 4.5$
Cosine starts at its maximum by default (no phase shift needed). The midline and amplitude are read directly from max/min values.
Cosine is symmetric about $\theta = 0$ (and $2\pi$), giving a second solution at $2\pi - \theta$ in $[0, 2\pi]$. A sketch of the cosine model confirms two intersections with $h = 5.25$ in one period.
PROBLEM 2 · DECIBEL INTENSITY COMPARISON
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?
Divide both sides by 10, then exponentiate: $10^{\text{(dB level}/10)} \times I_0$ gives $I$. This is the standard rearrangement of the decibel formula.
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Part (b), intensity ratio $\dfrac{I_{\text{café}}}{I_{\text{library}}} = \dfrac{10^{-5}}{10^{-9}} = 10^4 = 10\,000$
Alternatively using level difference: $\Delta L = 70 - 30 = 40$ dB ratio $= 10^{40/10} = 10^4 = 10\,000$
A 40 dB difference always corresponds to a $10^4$ intensity ratio, regardless of the reference. This shortcut ($10^{\Delta L / 10}$) is very useful in exam questions comparing two sound levels.
PROBLEM 3 · RICHTER SCALE, COMPARING EARTHQUAKES
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?
Part (b), interpreting the result $M_B - M_A = 6.70 - 5.0 = 1.70$ 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)
The log scale means even enormous physical differences appear as modest numerical differences. This is precisely why scientists prefer log scales, a magnitude-8 earthquake is catastrophically larger than magnitude 5, but the numbers still look manageable on paper.
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Retrieve and consolidate
Check the core ideas, then revisit the claims from the opening scenario.
Multiple choice · 5 questions
MC
Multiple choice review
+5 XP each
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:
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Revisit your thinking
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:
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Complete independent practice
Answer the lesson questions, show your working and compare your reasoning with the worked solutions.
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Short answer, exam-style questions
show all working
ApplyBand 44 marks
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)
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AnalyseBand 4–54 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)
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AnalyseBand 55 marks
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)
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EvaluateBand 64 marks
SA 4. A researcher has two sets of measurements to graph.
Set A, the apparent brightness of 12 stars, from $3 \times 10^{-13}$ to $1.4 \times 10^{-8}$ W/m². Set B, the heights of the 12 students in one class, from 152 cm to 187 cm.
(a) State which set should be plotted on a logarithmic scale, and justify your choice by referring to the range of each set. (2 marks)
(b) Explain what would be lost if a logarithmic scale were used for the other set. (1 mark)
(c) Stellar magnitude is defined as $m = -2.5\log_{10}(b/b_0)$, where $b$ is apparent brightness. Explain what the negative coefficient does to the order of the magnitudes. (1 mark)
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📖 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]
(c) The claim does not follow from the formula. The engineer has compared the dB numbers directly ($70 \approx 2 \times 40$), but the decibel scale is logarithmic, so dB values cannot be compared by multiplication. What the formula gives is the intensity ratio from part (b): 1000 times more intense. Perceived loudness is a claim about human hearing, not about $L$, so this formula cannot settle it either way. [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]
(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]
SA 4 (4 marks):
(a) Set A. Its values run from $3 \times 10^{-13}$ to $1.4 \times 10^{-8}$, a ratio of $\dfrac{1.4 \times 10^{-8}}{3 \times 10^{-13}} \approx 4.7 \times 10^{4}$, so the set spans about $\log_{10}(4.7 \times 10^{4}) \approx 4.7$ orders of magnitude. On a linear axis the faintest stars would be pressed against zero and indistinguishable from one another while the brightest set the scale. A logarithmic axis turns that ratio into a spread of about 4.7 units, so every star is separately readable. Set B runs from 152 cm to 187 cm, a ratio of only $187/152 \approx 1.23$. [2]
(b) Taking logs of Set B compresses it into $\log_{10}(152) \approx 2.18$ to $\log_{10}(187) \approx 2.27$, a spread of about 0.09 of a unit. The differences between students, which are the whole point of the graph, would be squeezed into almost nothing. A logarithmic scale also asks the reader to compare by ratio, and heights are compared by difference, so the axis would misrepresent what the data means as well as hide it. [1]
(c) The negative coefficient reverses the order. A larger brightness $b$ gives a larger $\log_{10}(b/b_0)$, which the $-2.5$ turns into a smaller (more negative) $m$. So the brightest objects carry the lowest magnitudes, which is why Sirius is about $-1.5$ and a star at the naked-eye limit is about $+6$. [1]
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Review and finish
Retrieve this lesson's modelling decisions, then extend your practice with the module quiz.
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Retrieve the modelling decisions
review
Without looking back, explain how maximum, minimum and period determine a periodic model. Then explain why equal steps on a base-10 logarithmic scale represent equal ratios, using one example from decibels, seismic magnitude, stellar magnitude or pH.
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.