Get Ready: Module 4 — Waves, Sound & Resonance
Everyone starts somewhere. Before we drive a resonance tube and chase harmonics, let's build the floor before the building — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.
Skills to Build First
Five small building blocks. Master these and Module 4 will feel like the next natural step, not a wall.
Ratios and Proportions
What it is & why you need it: A proportion sets two ratios equal, like \(\dfrac{a}{b} = \dfrac{c}{d}\). Waves live on proportions: the harmonic formula \(f_n = \dfrac{n v}{2L}\) says frequency is proportional to the mode number \(n\) and inversely proportional to the length \(L\). Sound intensity falls off as the inverse square of distance. If you can set up and solve a proportion, half of this module is bookkeeping.
Worked example: a tube's 1st harmonic is \(86\) Hz — find its 3rd
- The harmonics are proportional to \(n\): \(f_n = n\,f_1\). The 1st harmonic is \(f_1 = 86\) Hz.
- For the 3rd harmonic, set \(n = 3\): \(f_3 = 3 \times 86\).
- Multiply: \(f_3 = 258\) Hz.
- Read it as a ratio: \(\dfrac{f_3}{f_1} = \dfrac{3}{1}\) — three times the pitch, exactly as the proportion promised.
Inverse proportion flips the pattern: double the tube length \(L\) and the frequency \(f = \dfrac{v}{2L}\) is halved, not doubled.
1. A pipe's fundamental is \(120\) Hz. What is its 4th harmonic (open–open)?
2. A wave travels \(343\) m/s. If its frequency doubles, what happens to its wavelength (\(\lambda = v/f\))?
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2. Wavelength is inversely proportional to frequency. Doubling \(f\) halves \(\lambda\) — the speed \(v\) is fixed, so \(f\) and \(\lambda\) trade off.
Logarithms for Decibels
What it is & why you need it: A logarithm answers "ten to what power gives this number?" So \(\log(1000) = 3\) because \(10^3 = 1000\). Loudness is measured in decibels with \(\beta = 10\log\!\big(\tfrac{I}{I_0}\big)\), because our ears span a huge range of intensities. You only need base-10 logs of powers of ten — no calculator gymnastics.
Worked example: a sound is \(10^{6}\) times the reference intensity — find dB
- Form the ratio \(\dfrac{I}{I_0}\). Here it is \(10^{6}\) (a million times louder than the threshold).
- Take the base-10 log: \(\log(10^{6}) = 6\), because \(10^6 = 10^6\). (The log of a power of ten is just the exponent.)
- Multiply by \(10\): \(\beta = 10 \times 6 = 60\).
- So the sound level is \(60\) dB — about the loudness of a normal conversation.
Shortcut: \(\log(10^k) = k\). Every factor of \(10\) in intensity adds \(10\) dB; every factor of \(100\) adds \(20\) dB.
1. Evaluate \(\log(10^{4})\).
2. A sound has intensity \(10^{9}\) times \(I_0\). Find its level in decibels using \(\beta = 10\log\!\big(\tfrac{I}{I_0}\big)\).
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2. \(\beta = 10\log(10^{9}) = 10 \times 9 = 90\) dB (about as loud as heavy traffic).
Square Roots
What it is & why you need it: A square root undoes a square: \(\sqrt{49} = 7\) because \(7^2 = 49\). The speed of a wave on a stretched string is \(v = \sqrt{\dfrac{T}{\mu}}\) — tension over linear density, under a root. Tuning a guitar is literally adjusting a square root. You'll also meet roots whenever a formula has a squared quantity you need to peel back.
Worked example: a string has \(T = 100\) N and \(\mu = 0.0025\) kg/m — find \(v\)
- Write the formula: \(v = \sqrt{\dfrac{T}{\mu}}\).
- Substitute: \(v = \sqrt{\dfrac{100}{0.0025}}\).
- Divide inside the root first: \(\dfrac{100}{0.0025} = 40{,}000\).
- Take the root: \(v = \sqrt{40{,}000} = 200\) m/s.
Always simplify inside the root before taking it. And remember perfect squares: \(\sqrt{4}=2,\ \sqrt{9}=3,\ \sqrt{16}=4,\ \dots,\ \sqrt{144}=12\).
1. Evaluate \(\sqrt{81}\).
2. A string has \(T = 64\) N and \(\mu = 0.01\) kg/m. Find the wave speed \(v = \sqrt{T/\mu}\).
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2. \(\dfrac{64}{0.01} = 6400\), and \(\sqrt{6400} = 80\) m/s.
The Shape and Period of a Sinusoid
What it is & why you need it: A sinusoid is the smooth, repeating curve of \(\sin\) or \(\cos\). Its period is the length of one full cycle; its frequency is how many cycles fit in a second, and the two are reciprocals: \(f = \dfrac{1}{T}\). Every sound wave, every standing-wave snapshot in the lab, is a sinusoid — reading its period off the picture is step one.
Worked example: a wave repeats every \(0.004\) s — find its frequency
- The period is the time for one cycle: \(T = 0.004\) s.
- Frequency is the reciprocal of the period: \(f = \dfrac{1}{T}\).
- Substitute: \(f = \dfrac{1}{0.004}\).
- Compute: \(f = 250\) Hz. (Two hundred fifty cycles every second.)
Period and frequency are a seesaw: a short period means a high frequency, and vice versa. \(T\) is in seconds, \(f\) is in hertz (cycles per second).
1. A pendulum completes one swing every \(2\) s. What is its frequency?
2. A sound wave has frequency \(500\) Hz. What is its period?
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2. \(T = \dfrac{1}{f} = \dfrac{1}{500} = 0.002\) s \(= 2\) ms.
The Link to Simple Harmonic Motion (from Module 3)
What it is & why you need it: In Module 3 you met simple harmonic motion (SHM) — a mass on a spring or a swinging pendulum, oscillating with a definite period. A sound wave is SHM spread out in space: each tiny parcel of air does the same back-and-forth wiggle, just a beat out of step with its neighbor. So the period ↔ frequency relationship and the sinusoidal shape carry straight over from M3. Resonance is what happens when you drive that oscillation at its own natural frequency.
Worked example: a spring–mass oscillator — find its frequency from its period
- From M3, the period of a spring oscillator is \(T = 2\pi\sqrt{\dfrac{m}{k}}\). Suppose a setup gives \(T = 0.5\) s.
- The oscillation frequency is the reciprocal: \(f = \dfrac{1}{T}\).
- Substitute: \(f = \dfrac{1}{0.5} = 2\) Hz.
- This is the system's natural frequency — drive it at \(2\) Hz and you get resonance, the same idea the tube uses with air.
The bridge from M3 to M4: a vibrating object has a natural frequency \(f = 1/T\); resonance is matching a driving frequency to it. Strings, air columns, and springs all obey the same rule.
1. An oscillator has a period of \(0.25\) s. What is its natural frequency?
2. A tube has a natural (fundamental) frequency of \(150\) Hz. At what driving frequency will it resonate the strongest?
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2. Resonance happens when the driving frequency equals the natural frequency, so it rings loudest at \(150\) Hz.
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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A pipe's fundamental is \(90\) Hz. What is its 3rd harmonic (open–open)?
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Harmonics are proportional to \(n\): \(f_3 = 3 \times 90 = 270\) Hz. -
Evaluate \(\log(10^{5})\).
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\(\log(10^{5}) = 5\) — the exponent on the ten. -
A sound is \(10^{8}\) times the reference intensity. Find its level using \(\beta = 10\log\!\big(\tfrac{I}{I_0}\big)\).
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\(\beta = 10 \times 8 = 80\) dB. -
Evaluate \(\sqrt{225}\).
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\(\sqrt{225} = 15\), since \(15^2 = 225\). -
A wave has period \(0.005\) s. What is its frequency?
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\(f = \dfrac{1}{0.005} = 200\) Hz. -
An oscillator has a natural frequency of \(180\) Hz. At what driving frequency does it resonate strongest?
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At its natural frequency: \(180\) Hz. Resonance means driving frequency = natural frequency.
If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Ratios & proportions
Logarithms (decibels)
Square roots
Period & frequency of waves
Simple harmonic motion
When You're Ready →
No rush. When the skills above feel solid, step into Module 4.
Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Beta Academy
AP Physics · enrichment — a supplemental, college-level course offered alongside the Mathematical Architects sequence.
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