High school · ScienceAbout 180 minutes
ScienceHigh schoolAbout 180 minutes

Apply conservation of energy, reflection, refraction, diffraction and interference to waves, and solve standing-wave problems

Several sessions: a laboratory investigation with low-voltage or mechanical apparatus only, modeled problems checked in SI units, independent practice and a written explanation. g = 9.8 m/s² throughout.

Ohio learning standards this draft targets

science-2018:P.W.1

Read the standard text

science-2018:P.W.1 Wave properties

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 · LearnLearn the idea
The string is fixed at 0 and 1.20 m. In the second harmonic the wavelength equals the string length, 1.20 m, so with v = 60 m/s the frequency is 60 ÷ 1.20 = 50 Hz. Nodes, which never move, are at 0, 0.60 and 1.20 m; antinodes with amplitude 0.10 m (10 cm) are at 0.30 and 0.90 m. The two curves show the string at one instant of greatest displacement and half a period (0.010 s) later, when every point is displaced the opposite way.The string is fixed at 0 and 1.20 m. In the second harmonic the wavelength equals the string length, 1.20 m, so with v = 60 m/s the frequency is 60 ÷ 1.20 = 50 Hz. Nodes, which never move, are at 0, 0.60 and 1.20 m; antinodes with amplitude 0.10 m (10 cm) are at 0.30 and 0.90 m. The two curves show the string at one instant of greatest displacement and half a period (0.010 s) later, when every point is displaced the opposite way.Second harmonic on a 1.20 m string (v = 60m/s, f = 50 Hz)00.20.40.60.811.2-12-8-404812nodenodenodeantinodeantinodeposition along string (m)displacement (cm)
The string is fixed at 0 and 1.20 m. In the second harmonic the wavelength equals the string length, 1.20 m, so with v = 60 m/s the frequency is 60 ÷ 1.20 = 50 Hz. Nodes, which never move, are at 0, 0.60 and 1.20 m; antinodes with amplitude 0.10 m (10 cm) are at 0.30 and 0.90 m. The two curves show the string at one instant of greatest displacement and half a period (0.010 s) later, when every point is displaced the opposite way.

Learning goal: Use wavelength, frequency, amplitude and speed to describe waves quantitatively, account for the energy of a wave at a boundary, apply the law of reflection and describe refraction and diffraction of waves, locate constructive and destructive interference from path differences, and calculate the frequencies of standing waves on strings and in open and closed pipes.

Measurable properties. A wave is a traveling disturbance that carries energy from one place to another without carrying the medium with it. Its wavelength λ is the distance between successive crests, its frequency f the number of complete cycles that pass a point each second (hertz, Hz), its period T = 1/f, and its amplitude A the maximum displacement from equilibrium. The speed of a wave is set by the medium, not by the source (for the springs, strings and sound in air used here, it is also the same for all frequencies; in some media, such as water waves in deep water, it is not): a string's tension and mass per length, the depth of water, the temperature of air (about 343 m/s for sound at 20 °C). Speed, frequency and wavelength are linked by v = fλ. For waves on a string or in air the energy carried is proportional to the square of the amplitude, so, of two otherwise identical waves (same frequency, same medium), the one with twice the amplitude carries four times the energy. Worked model: a wave travels along a rope at 40 m/s while the hand driving it oscillates at 50 Hz; λ = v/f = 40/50 = 0.80 m.

Energy at a boundary is conserved. Follow the energy of a wave that arrives at a wall, or at the join between one medium and the next. It splits. One share bounces back as a reflected wave. Another share crosses the boundary: some of that keeps going as a transmitted wave in the second medium, and some is absorbed by the material and ends up in other forms, mostly thermal energy. A further share can leave the wave altogether, warming the boundary itself or being passed to nearby objects that the wave sets moving. These splits generally happen together, and the shares must add up to what arrived: incident energy = reflected wave energy + transmitted wave energy + energy converted to other forms (mostly thermal energy of the material). A pulse on a light, low-mass rope tied to a heavier rope reflects partly and transmits partly; a sound wave meeting a wall is partly reflected as an echo, partly transmitted through the wall, and partly absorbed. If a boundary reflects 30 % of a wave's energy and absorbs 20 %, then 50 % continues as a wave in the new medium.

Reflection and refraction of waves. When a wave bounces off a barrier, the angle at which it approaches (the angle of incidence, measured from the normal to the barrier) equals the angle at which it leaves (the angle of reflection). When a wave traveling through a two-dimensional medium (surface water, seismic waves) or a three-dimensional one (sound, electromagnetic waves) enters a medium where it travels at a different speed, both its speed and its wavelength change, while its frequency stays fixed by the source. If the wave meets the boundary at an angle, its direction also changes: this bending is refraction. Water waves slow in shallower water, so a wave crossing from deep to shallow water at 4.0 Hz with speed changing from 2.0 m/s to 1.2 m/s keeps its frequency while its wavelength shrinks from 0.50 m to 0.30 m, and if it arrived at an angle its direction of travel turns toward the normal (its crests, which are perpendicular to that direction, turn toward being parallel to the boundary).

Diffraction. Waves bend around barriers and spread out after passing through openings. The amount of bending increases as the wavelength increases relative to the size of the obstacle or opening; when the wavelength is much smaller than the opening, no noticeable diffraction occurs. This is why you hear someone around a corner (sound wavelengths are meters) but cannot see them (light wavelengths are less than a micrometer). Passing waves through a single slit or a double slit produces diffraction patterns of alternating constructive and destructive interference, and those patterns change predictably: narrowing a slit spreads the pattern out, increasing the wavelength spreads it out, and increasing the spacing between two slits brings the bright regions closer together. (The calculations for light are in the light-phenomena lessons; the wavelet model of how a wavefront propagates is not used in this course.)

Superposition and interference. When two waves occupy the same place at the same time, the displacement is the sum of the individual displacements; afterwards each wave continues unchanged. Two crests add to a larger crest (constructive interference); a crest and an equal trough cancel (destructive interference). For two sources vibrating in step, the result at any point depends on the path difference, the difference between the distances from the two sources to that point. A path difference of a whole number of wavelengths (0, λ, 2λ, ...) gives constructive interference; an odd number of half wavelengths (½λ, 1½λ, ...) gives destructive interference. Marking these points across a region produces the two-source pattern of loud and quiet lines (for sound) or bright and dark fringes (for light). Worked model: two speakers in step emit a 686 Hz tone (λ = 343/686 = 0.50 m). A listener is 3.0 m from one speaker and 3.5 m from the other. Path difference 0.50 m = 1λ, so the sound is loud there. If the frequency drops to 343 Hz (λ = 1.0 m), the same 0.50 m is ½λ and the sound is faint.

Standing waves. A wave reflected from a fixed end travels back through the incoming wave. At certain frequencies the two interfere to form a standing wave: a pattern that oscillates in place with nodes, points that never move, and antinodes of maximum motion between them. The condition is that the length of the medium fits the wave: for a string fixed at both ends (or an air column open at both ends), both ends are nodes (for the string) or antinodes (for the open pipe), and the allowed wavelengths are λₙ = 2L/n with n = 1, 2, 3, ..., giving harmonics fₙ = nv/(2L). For a pipe closed at one end and open at the other, the closed end is a node and the open end an antinode, so λₙ = 4L/n with n odd only, fₙ = nv/(4L), n = 1, 3, 5, .... In a pipe these nodes and antinodes describe the back-and-forth displacement of the air: the air cannot move into the closed end (a displacement node) and moves most freely at an open end (a displacement antinode). The pressure variation is the other way round, largest at a closed end and smallest at an open end. The formulas also place the antinode exactly at the open end; in a real pipe it sits slightly beyond the end, so measured frequencies come out a little lower than predicted (an end correction that this course does not calculate). The lowest frequency is the fundamental; higher harmonics are whole-number multiples of it (odd multiples only for a closed pipe). Adjacent nodes are half a wavelength apart, which is the quickest way to read λ from a pattern. Worked model: a 1.20 m string carries waves at 60 m/s. Fundamental: λ₁ = 2.40 m, f₁ = 60/2.40 = 25 Hz. Second harmonic: λ₂ = 1.20 m, f₂ = 50 Hz, with nodes at 0, 0.60 and 1.20 m and antinodes at 0.30 and 0.90 m, as the diagram shows. Third harmonic: 75 Hz.

Misconception: a standing wave is not moving, so no energy is involved. The string is oscillating everywhere except at the nodes; energy is stored in the motion and stretch of the string and no net energy travels along it because the two traveling waves carry equal energy in opposite directions. Energy still enters from the driver and leaves through friction and sound, which is why a plucked string eventually falls silent.

Investigation: standing waves on a cord. Materials: 3 m of elastic cord or a long spring, a stopwatch, a meter stick. Steps: one student holds an end fixed; the other oscillates the far end by hand and adjusts the frequency until a stable one-loop pattern forms, then two loops, then three. For each, time ten oscillations to find f, measure L and count loops to find λ = 2L/n, and calculate v = fλ. A result that supports the model: v is the same for all three patterns (within about 10 %, a suggested classroom criterion to be judged against your own timing and length uncertainties), and the frequencies are in the ratio 1 : 2 : 3. What the model cannot show: the driven end is not a true node, and energy lost to the air is not represented, so real loops decay when the driver stops. Safety: keep the cord well below head height and never release a stretched cord.

Section 2 · ModelWork through the examples

Worked example 1 — speed, frequency and wavelength. Sound at 343 m/s with a frequency of 686 Hz has λ = 343/686 = 0.50 m. Raising the frequency to 1372 Hz halves the wavelength to 0.25 m; the speed does not change because it depends only on the air.

Worked example 2 — harmonics on a string. A 0.80 m guitar string carries waves at 120 m/s. f₁ = v/(2L) = 120/1.60 = 75 Hz; f₂ = 150 Hz; f₃ = 225 Hz. Nodes for the third harmonic are at 0, 0.267, 0.533 and 0.80 m, one third of a length apart, and adjacent nodes are half a wavelength (λ₃ = 0.533 m) apart.

Worked example 3 — open and closed pipes. A pipe 0.50 m long with air at 343 m/s. Open at both ends: f₁ = v/(2L) = 343/1.00 = 343 Hz, then 686 Hz, 1029 Hz. Closed at one end: f₁ = v/(4L) = 343/2.00 = 171.5 Hz, and only odd harmonics follow: 514.5 Hz, 857.5 Hz. Closing one end halves the fundamental and removes the even harmonics, because the closed end must be a displacement node and the open end a displacement antinode, so only an odd number of quarter wavelengths fits (end corrections neglected).

Worked example 4 — two-source interference. Two in-step sources emit waves of wavelength 0.40 m. Point P is 2.6 m from one source and 3.4 m from the other: path difference 0.80 m = 2λ, constructive. Point Q is 2.6 m and 3.2 m away: difference 0.60 m = 1½λ, destructive. Drawing the two sets of circular crests and marking where crest meets crest (constructive) and crest meets trough (destructive) gives the same lines.

Worked example 5 — energy at a boundary. A wave pulse carrying 12 J of energy meets a boundary that reflects 25 % and absorbs 15 %. Reflected: 3.0 J; absorbed (to thermal energy): 1.8 J; transmitted as a wave: 12 − 3.0 − 1.8 = 7.2 J. The account balances because energy is conserved.

Section 3 · DoPractice, demonstrate, and explain

Representation task: draw and label patterns. Draw the first three standing-wave patterns for a 1.20 m string (nodes and antinodes labeled with positions in meters) and the first two for a 0.50 m closed pipe, stating the wavelength and frequency of each for v = 60 m/s (string) and 343 m/s (pipe). Then sketch two sets of circular crests from in-step sources 1.0 m apart with λ = 0.40 m and mark at least three points of constructive and three of destructive interference, giving the path difference at each.

Independent performance: a standing-wave investigation report. Carry out the cord investigation for at least three harmonics and write a report. Success criteria: a table of loops, length, timed oscillations, frequency, wavelength and calculated speed in SI units; a check that the frequencies are close to the ratio 1 : 2 : 3 and that the speed is constant within a stated percentage; an explanation of the conditions required for a standing wave to form (reflection at the ends and a length that fits a whole number of half wavelengths); a calculation of the frequency of the fifth harmonic from your measured speed; a paragraph applying conservation of energy to what happened to the wave energy when you stopped driving the cord; and one statement of what the ideal model could not show in your set-up.

Section 4 · Check7-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

A wave on a lake has a frequency of 0.40 Hz and crests 3.5 m apart. Calculate its speed, and explain what would change if the same waves moved into shallower water where they travel more slowly.

Show worked solution

v = fλ = (0.40 Hz)(3.5 m) = 1.4 m/s. In shallower water the speed drops; the frequency stays 0.40 Hz because it is set by the source, so the wavelength shortens (λ = v/f). If the waves meet the boundary at an angle, their direction also changes, bending toward the normal: that is refraction.

Problem 2

A sound wave carrying 20 J of energy strikes a wall that reflects 35 % and absorbs 45 % of it. How much energy continues as a wave through the wall, and why must these fractions add to 100 %?

Show worked solution

Reflected: 7.0 J; absorbed and transformed to thermal energy: 9.0 J; transmitted through the wall as a wave: 20 − 7.0 − 9.0 = 4.0 J (20 %). Energy is conserved: everything that arrives at the boundary must be reflected, absorbed, transmitted or dissipated to the surroundings, so the parts must add to the incident energy.

Problem 3

Explain why you can hear a conversation around the corner of a building but cannot see the people talking, using the idea of diffraction.

Show worked solution

Diffraction, the bending of waves around obstacles and through openings, is noticeable only when the wavelength is comparable to or larger than the obstacle or opening. Audible sound has wavelengths from centimeters to meters, similar to the building corner and doorway, so it bends around the corner. Visible light has wavelengths of less than a micrometer, far smaller than the corner, so it shows no noticeable diffraction and travels in straight lines past it.

Problem 4

Two in-step speakers emit a 500 Hz tone (take v = 343 m/s). A listener is 4.20 m from one speaker and 4.89 m from the other. Is the sound loud or faint there? Show the calculation.

Show worked solution

λ = 343/500 = 0.686 m. Path difference = 4.89 − 4.20 = 0.69 m ≈ 1.0λ. A path difference of a whole number of wavelengths means the two waves arrive in step, crest with crest, so they interfere constructively and the sound is loud. Moving to a point where the difference is 0.343 m (half a wavelength) would make it faint.

Problem 5

A string 0.60 m long fixed at both ends has a fundamental frequency of 110 Hz. Find the wave speed on the string and the frequency of its third harmonic, and describe where the nodes of the third harmonic are.

Show worked solution

For the fundamental, λ₁ = 2L = 1.20 m, so v = fλ = (110)(1.20) = 132 m/s. The third harmonic has f₃ = 3f₁ = 330 Hz (λ₃ = 0.40 m). Its nodes are at 0, 0.20, 0.40 and 0.60 m: adjacent nodes are half a wavelength (0.20 m) apart, with antinodes midway between them.

Problem 6

A pipe closed at one end is 0.35 m long (v = 343 m/s). Calculate its fundamental frequency and the next harmonic it can produce, and explain why the second harmonic (2f₁) is missing.

Show worked solution

Closed pipe: λ₁ = 4L = 1.40 m, f₁ = 343/1.40 = 245 Hz. The next harmonic is the third, f₃ = 3 × 245 = 735 Hz. The closed end must be a node of the air's displacement and the open end an antinode of it, so the pipe length must equal an odd number of quarter wavelengths (L = λ/4, 3λ/4, 5λ/4, ...). A wavelength giving 2f₁ would need L = λ/2, which puts the same kind of point (two nodes or two antinodes) at both ends, so that pattern cannot form. (These values neglect the small end correction of a real pipe.)

Problem 7

State the conditions required for a standing wave to form on a string and explain what happens to the energy of a standing wave after the driver stops.

Show worked solution

Two conditions: the wave must be reflected at the ends so that identical waves travel in opposite directions through each other, and the string's length must fit a whole number of half wavelengths so that the reflected wave reinforces the incoming one (fixed ends are nodes). The standing wave stores energy in the string's motion and stretch, with no net energy transport along the string. After the driver stops, no energy enters, while friction, air resistance and sound carry energy out, so the amplitude decays until the string is still; the total energy is conserved as thermal energy and sound in the surroundings.