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AP Physics · Foundations

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.

Feeling shaky? That's completely okay — it just means you're about to fill in a gap that will make the whole module easier. Read the plain-language explanation, follow the worked example step by step, then try one yourself. The answers are hidden so you can think first — click Show answer whenever you're ready. You can do this.

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Skills to Build First

Five small building blocks. Master these and Module 4 will feel like the next natural step, not a wall.

Skill 01

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

  1. The harmonics are proportional to \(n\): \(f_n = n\,f_1\). The 1st harmonic is \(f_1 = 86\) Hz.
  2. For the 3rd harmonic, set \(n = 3\): \(f_3 = 3 \times 86\).
  3. Multiply: \(f_3 = 258\) Hz.
  4. 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.

Try it

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 (\(\lambda = v/f\))?

Show answer
1. Harmonics are proportional to \(n\): \(f_4 = 4 \times 120 = 480\) Hz.

2. Wavelength is inversely proportional to frequency. Doubling \(f\) halves \(\lambda\) — the speed \(v\) is fixed, so \(f\) and \(\lambda\) trade off.
Skill 02

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

  1. Form the ratio \(\dfrac{I}{I_0}\). Here it is \(10^{6}\) (a million times louder than the threshold).
  2. 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.)
  3. Multiply by \(10\): \(\beta = 10 \times 6 = 60\).
  4. 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.

Try it

1. Evaluate \(\log(10^{4})\).

2. A sound has intensity \(10^{9}\) times \(I_0\). Find its level in decibels using \(\beta = 10\log\!\big(\tfrac{I}{I_0}\big)\).

Show answer
1. \(\log(10^{4}) = 4\) — the exponent on the ten.

2. \(\beta = 10\log(10^{9}) = 10 \times 9 = 90\) dB (about as loud as heavy traffic).
Skill 03

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 \(v\)

  1. Write the formula: \(v = \sqrt{\dfrac{T}{\mu}}\).
  2. Substitute: \(v = \sqrt{\dfrac{100}{0.0025}}\).
  3. Divide inside the root first: \(\dfrac{100}{0.0025} = 40{,}000\).
  4. Take the root: \(v = \sqrt{40{,}000} = 200\) m/s.

Always simplify inside the root before taking it. And remember perfect squares: \(\sqrt{4}=2,\ \sqrt{9}=3,\ \sqrt{16}=4,\ \dots,\ \sqrt{144}=12\).

Try it

1. Evaluate \(\sqrt{81}\).

2. A string has \(T = 64\) N and \(\mu = 0.01\) kg/m. Find the wave speed \(v = \sqrt{T/\mu}\).

Show answer
1. \(\sqrt{81} = 9\), since \(9^2 = 81\).

2. \(\dfrac{64}{0.01} = 6400\), and \(\sqrt{6400} = 80\) m/s.
Skill 04

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

  1. The period is the time for one cycle: \(T = 0.004\) s.
  2. Frequency is the reciprocal of the period: \(f = \dfrac{1}{T}\).
  3. Substitute: \(f = \dfrac{1}{0.004}\).
  4. 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).

Try it

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?

Show answer
1. \(f = \dfrac{1}{T} = \dfrac{1}{2} = 0.5\) Hz — half a cycle per second.

2. \(T = \dfrac{1}{f} = \dfrac{1}{500} = 0.002\) s \(= 2\) ms.
Skill 05

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

  1. From M3, the period of a spring oscillator is \(T = 2\pi\sqrt{\dfrac{m}{k}}\). Suppose a setup gives \(T = 0.5\) s.
  2. The oscillation frequency is the reciprocal: \(f = \dfrac{1}{T}\).
  3. Substitute: \(f = \dfrac{1}{0.5} = 2\) Hz.
  4. 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.

Try it

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?

Show answer
1. \(f = \dfrac{1}{0.25} = 4\) Hz.

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.

  1. A pipe's fundamental is \(90\) Hz. What is its 3rd harmonic (open–open)?
    Show answer
    Harmonics are proportional to \(n\): \(f_3 = 3 \times 90 = 270\) Hz.
  2. Evaluate \(\log(10^{5})\).
    Show answer
    \(\log(10^{5}) = 5\) — the exponent on the ten.
  3. A sound is \(10^{8}\) times the reference intensity. Find its level using \(\beta = 10\log\!\big(\tfrac{I}{I_0}\big)\).
    Show answer
    \(\beta = 10 \times 8 = 80\) dB.
  4. Evaluate \(\sqrt{225}\).
    Show answer
    \(\sqrt{225} = 15\), since \(15^2 = 225\).
  5. A wave has period \(0.005\) s. What is its frequency?
    Show answer
    \(f = \dfrac{1}{0.005} = 200\) Hz.
  6. An oscillator has a natural frequency of \(180\) Hz. At what driving frequency does it resonate strongest?
    Show answer
    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.


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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


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When You're Ready →

No rush. When the skills above feel solid, step into Module 4.

Module 4 Visual Lab

Drive the resonance tube and see standing waves lock in when you hit a harmonic.

Go to Module 4 →

Module 3 Refresher

Simple harmonic motion is the seed of every wave. Revisit the springs and pendulums it grows from.

Open Module 3

AP Physics Overview

The full picture of this self-study enrichment track — five modules, no syllabus, no pressure.

Back to AP Physics

Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Beta Academy
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