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Physics · enrichment · Foundations

Get Ready: Module 4 — Waves, Sound & Optics

Everyone starts somewhere. Before we ripple a tank and bend some light, 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. This is enrichment — not a TEKS-graded course, so explore freely.
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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

Reciprocals: \(\frac{1}{T}\) and \(\frac{1}{f}\)

What it is & why you need it: The reciprocal of a number is just \(1\) divided by it — flip the fraction over. In Module 4 frequency and period are reciprocals: \(f = \dfrac{1}{T}\) and \(T = \dfrac{1}{f}\). If a wave repeats every \(0.25\) s, it happens \(4\) times per second. That flip is one of the most-used moves all module.

Worked example: a wave has period \(T = 0.25\) s — find its frequency

  1. Write the relationship: frequency is the reciprocal of period, \(f = \dfrac{1}{T}\).
  2. Substitute the period: \(f = \dfrac{1}{0.25}\).
  3. Divide. Dividing by \(0.25\) is the same as multiplying by \(4\): \(\dfrac{1}{0.25} = 4\).
  4. So \(f = 4\) Hz — the wave completes \(4\) cycles every second.

Quick check: \(f\) and \(T\) always multiply to \(1\). Here \(f \times T = 4 \times 0.25 = 1\). \(\checkmark\)

Try it

1. A pendulum swings with period \(T = 0.5\) s. What is its frequency?

2. A tuning fork vibrates at \(f = 50\) Hz. What is its period \(T\)?

Show answer
1. \(f = \dfrac{1}{T} = \dfrac{1}{0.5} = 2\) Hz.

2. \(T = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\) s.
Skill 02

\(\sin\) and Inverse \(\sin^{-1}\) on a Calculator

What it is & why you need it: \(\sin\theta\) takes an angle and gives a ratio between \(-1\) and \(1\); \(\sin^{-1}\) (inverse sine, the asin key) runs it backward — you give the ratio, it returns the angle. Snell's law and the critical-angle problems in Module 4 need both directions. Make sure your calculator is in DEGREE mode for these.

Worked example: find \(\sin 30^\circ\), then find the angle whose sine is \(0.5\)

  1. Set the calculator to degree mode (look for DEG on the display, not RAD).
  2. Forward: type sin(30). The result is \(0.5\). So \(\sin 30^\circ = 0.5\).
  3. Backward: type sin⁻¹(0.5) (often 2nd then sin). The result is \(30^\circ\).
  4. They undo each other: \(\sin^{-1}(\sin 30^\circ) = 30^\circ\).

Tip: if you get a weird decimal like \(-0.988\) for \(\sin 30\), your calculator is in radian mode — switch to degrees and try again.

Try it

1. Use your calculator (degree mode) to find \(\sin 40^\circ\), rounded to 3 decimals.

2. Find the angle \(\theta\) (in degrees) whose sine is \(0.4285\), rounded to the nearest tenth.

Show answer
1. \(\sin 40^\circ \approx 0.643\).

2. \(\theta = \sin^{-1}(0.4285) \approx 25.4^\circ\). (These are exactly the numbers in the module's Snell's-law example.)
Skill 03

Recall the Constants: \(c = 3\times10^8\) m/s and \(v_{\text{sound}} \approx 343\) m/s

What it is & why you need it: Two speeds anchor this whole module. The speed of light in vacuum is \(c = 3\times10^{8}\) m/s (300 million meters every second). The speed of sound in air is about \(v_{\text{sound}} \approx 343\) m/s at room temperature. You'll plug these into \(v = f\lambda\), the Doppler formula, and \(n = \dfrac{c}{v}\) constantly — knowing them cold saves you every time.

Worked example: a 686 Hz sound in air — find its wavelength

  1. Recall the speed of sound: \(v_{\text{sound}} \approx 343\) m/s.
  2. Rearrange the wave equation for wavelength: \(\lambda = \dfrac{v}{f}\).
  3. Substitute: \(\lambda = \dfrac{343}{686}\).
  4. Divide: \(\dfrac{343}{686} = 0.5\) m. (Notice \(686 = 2 \times 343\), so the answer is exactly one-half.)

Scale sense: light is almost a million times faster than sound — that's why you see lightning long before you hear the thunder.

Try it

1. Light travels at \(2\times10^{8}\) m/s in glass. Using \(c = 3\times10^{8}\) m/s, find the index of refraction \(n = \dfrac{c}{v}\).

2. About how long does sound take to travel \(686\) m through air? (Use \(v_{\text{sound}} \approx 343\) m/s and \(t = \dfrac{d}{v}\).)

Show answer
1. \(n = \dfrac{c}{v} = \dfrac{3\times10^{8}}{2\times10^{8}} = 1.5\).

2. \(t = \dfrac{d}{v} = \dfrac{686}{343} = 2\) s.
Skill 04

Reading Amplitude and Wavelength off a Sinusoid

What it is & why you need it: A wave drawn on a graph is a sine curve. Its amplitude is the height from the middle line to a crest (how tall), and its wavelength \(\lambda\) is the horizontal distance from one crest to the next (how long). Reading these straight off a picture is the first thing you do with any wave graph in Module 4.

Worked example: a wave graph rises to \(+3\) cm and dips to \(-3\) cm, with crests \(2\) m apart

  1. Find the midline. It sits halfway between the top (\(+3\)) and bottom (\(-3\)) — the midline is \(0\).
  2. Amplitude = crest height above the midline. From \(0\) up to \(+3\) is \(3\) cm, so amplitude \(A = 3\) cm.
  3. Wavelength = crest-to-crest distance. Adjacent crests are \(2\) m apart, so \(\lambda = 2\) m.
  4. Done — \(A = 3\) cm tall, \(\lambda = 2\) m long. (Amplitude is half the full top-to-bottom height of \(6\) cm.)

Watch the trap: amplitude is measured from the middle to a crest, not from the trough all the way up to the crest. The full peak-to-trough swing is \(2A\).

Try it

1. A wave swings from a crest at \(+8\) units to a trough at \(-8\) units. What is its amplitude?

2. On a graph, one crest is at \(x = 1\) m and the very next crest is at \(x = 4\) m. What is the wavelength?

Show answer
1. Amplitude is from the midline (\(0\)) to a crest, so \(A = 8\) units. (The full swing is \(16\), which is \(2A\).)

2. Crest-to-crest is \(4 - 1 = 3\) m, so \(\lambda = 3\) m.
Skill 05

Cross-Multiplying a Proportion

What it is & why you need it: A proportion sets two fractions equal, like \(\dfrac{a}{b} = \dfrac{c}{d}\). To solve for a missing piece you cross-multiply: \(a \cdot d = b \cdot c\). Snell's law \(n_1\sin\theta_1 = n_2\sin\theta_2\) and the wave equation are both proportions in disguise — this one algebra move unlocks a huge share of Module 4.

Worked example: solve \(\dfrac{3}{4} = \dfrac{x}{8}\) for \(x\)

  1. Cross-multiply. Multiply each top by the other bottom: \(3 \cdot 8 = 4 \cdot x\).
  2. Simplify the known side: \(24 = 4x\).
  3. Divide to isolate \(x\): \(x = \dfrac{24}{4}\).
  4. Simplify: \(x = 6\). Check: \(\dfrac{6}{8} = \dfrac{3}{4}\). \(\checkmark\)

Why it works: cross-multiplying is just multiplying both sides by both denominators at once — it clears the fractions in one stroke.

Try it

1. Solve \(\dfrac{x}{5} = \dfrac{12}{4}\) for \(x\).

2. A Snell's-law step gives \(1.0 \cdot 0.6 = 1.5 \cdot \sin\theta_2\). Solve for \(\sin\theta_2\).

Show answer
1. Cross-multiply: \(4x = 5 \cdot 12 = 60\). Divide by 4: \(x = 15\).

2. Divide both sides by \(1.5\): \(\sin\theta_2 = \dfrac{1.0 \cdot 0.6}{1.5} = \dfrac{0.6}{1.5} = 0.4\).
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Quick Readiness Check

Six short questions across all five skills. Try each one first, then peek at the answer.

  1. A wave has period \(T = 0.2\) s. Find its frequency \(f\).
    Show answer
    \(f = \dfrac{1}{T} = \dfrac{1}{0.2} = 5\) Hz.
  2. In degree mode, what is \(\sin 30^\circ\)? And \(\sin^{-1}(0.5)\)?
    Show answer
    \(\sin 30^\circ = 0.5\), and \(\sin^{-1}(0.5) = 30^\circ\). The two operations undo each other.
  3. What is the approximate speed of sound in air, and the speed of light in vacuum?
    Show answer
    Sound: \(v_{\text{sound}} \approx 343\) m/s. Light: \(c = 3\times10^{8}\) m/s.
  4. A wave swings between \(+5\) cm and \(-5\) cm. What is its amplitude?
    Show answer
    Amplitude is midline-to-crest: \(A = 5\) cm. (The full swing of \(10\) cm equals \(2A\).)
  5. Crests appear at \(x = 2\) m and \(x = 6\) m on a wave graph. What is the wavelength?
    Show answer
    \(\lambda = 6 - 2 = 4\) m (crest-to-crest distance).
  6. Solve the proportion \(\dfrac{x}{6} = \dfrac{10}{3}\).
    Show answer
    Cross-multiply: \(3x = 60\), so \(x = 20\).

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.)

Reciprocals (\(1/T\), \(1/f\))

Khan Academy IXL

Sine & inverse sine

Khan Academy IXL

Scientific notation & constants

Khan Academy IXL

Reading sine graphs

Khan Academy IXL

Proportions & cross-multiplying

Khan Academy IXL
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When You're Ready →

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

Module 4 Overview

The full picture of Waves, Sound & Optics — vocabulary, worked examples, and the interactive lab.

Go to Module 4 →

The Wave Interference Lab

Place two sources, sweep frequency and wavelength, and see constructive and destructive bands come to life.

Open the Visual Lab

Student Support

Office-hours, tutoring, and how to get unstuck across every Studio course.

Student Support

Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Dr. Ijezie's STEM Studio
Physics enrichment foundations — offered as exploratory material alongside the graded mathematics pathway. This is not a TEKS-graded course. Constants used: \(v_{\text{sound}} \approx 343\) m/s, \(c = 3\times10^{8}\) m/s.