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Waves · Wavelength & Energy

Wavelength and Energy of a Wave

Drive a string into a standing wave, measure its wavelength with the on-screen ruler, and confirm it against λ = 2L/n. Then change the amplitude and frequency and calculate the energy the wave carries with K = ¼ μA²ω²λ — watching the energy climb with the square of both amplitude and frequency.

Theory — Wavelength and Wave Energy

Waves Carry Energy, Not Matter

A wave transfers energy through the oscillation of a medium without transporting the medium itself. On a string, each piece moves up and down while the wave pattern travels along — or, for a standing wave, stays fixed in place with stationary nodes and antinodes.

Wavelength of a Standing Wave

The wavelength λ is the distance between two consecutive crests or troughs — equivalently, twice the distance between adjacent nodes. For a string of length L fixed at both ends, only certain wavelengths "fit," set by the number of segments n.

Standing-Wave Wavelength λ = 2L / n

L = length of the string (m)
n = number of segments (loops) along the string
v = f · λ  (wave speed relates frequency and wavelength)
Only wavelengths λ = 2L/n produce a stable standing wave

Energy of a Wave

The energy a wave carries depends on the amplitude, the angular frequency, the wavelength, and the medium's linear mass density. The energy rises with the square of both the amplitude and the frequency.

Wave Energy K = ¼ · μ · A² · ω² · λ

μ = linear mass density (kg/m)
A = amplitude (m)
ω = angular frequency = 2πf (rad/s)
λ = wavelength (m)
K ∝ A² and K ∝ ω² (∝ f²) — double either and the energy quadruples

Amplitude

K ∝ A². Doubling the amplitude multiplies the wave energy by four.

Frequency

K ∝ ω² = (2πf)². Doubling the frequency multiplies the energy by four — and shortens the wavelength.

Wavelength

λ = 2L/n. More segments → shorter wavelength → higher frequency for the same wave speed.

QuantityRelationshipNotes
Wavelengthλ = 2L/nn = number of segments
Angular frequencyω = 2πfrad/s
Wave speedv = f·λset by tension & density
Wave energyK = ¼μA²ω²λ∝ A², ∝ f²

Instructions — Running the Virtual Experiment

The Wavelength tab drives the string into a standing wave so you can measure λ with the ruler and check it against 2L/n; the Wave Energy tab lets you set amplitude and frequency and computes K. Record every reading in your lab report with screenshots.

Part 1 — Measure the Wavelength (Wavelength tab)
1
Open Simulation → Wavelength. Set a frequency to drive the string into a clean standing wave with whole loops. Use the on-screen ruler to read the measured wavelength λₑ (crest to crest, or twice the node spacing).
2
Count the number of segments n and compute the theoretical wavelength λₜ = 2L/n. Compare λₑ with λₜ.
Part 2 — Calculate the Energy (Wave Energy tab)
1
Open Wave Energy. Set the amplitude A and frequency f for each scenario (μ = 0.0015 kg/m). The readout gives ω = 2πf and the wave energy K = ¼μA²ω²λ for both the measured and theoretical wavelengths.
2
Record A, f, ω, λ, and K for each scenario. Note how K changes when you increase the amplitude or the frequency.

Simulation — Wave on a String

Wavelength & Energy Virtual LabMeasure λ, then compute the wave energy
string
ruler (crest spacing)
String length L = 2.0 m · wave speed v = 12 m/s.

Drive the string

Wavelength readout
Segments n2
λ measured (ruler)2.00 m
λ theory = 2L/n2.00 m
Frequency f = v/λ6.00 Hz
λ = 2L/n. Measured and theoretical agree.
μ = 0.0015 kg/m · L = 2.0 m · amplitude exaggerated for display.

Amplitude & frequency

Energy readout
Amplitude A0.0050 m
ω = 2πf7.85 rad/s
λ = 2L/n2.00 m
μ0.0015 kg/m
K = ¼μA²ω²λ— J
K rises with A² and f².
ScenarioA (m)f (Hz)ω (rad/s)λ (m)μ (kg/m)K (J)
Set A, f, n and click "Record" to log a row.

Team Questions

Question 1. A string of length L = 2.0 m vibrates in n = 2 segments. Find the theoretical wavelength using λ = 2L/n. (Type in m — e.g. 2.0)
Question 2. For a wave of frequency f = 1.25 Hz, find the angular frequency ω = 2πf. (Type in rad/s to 3 sig figs — e.g. 7.85)
Question 3. Using K = ¼μA²ω²λ with μ = 0.0015 kg/m, A = 0.0050 m, ω = 7.85 rad/s, λ = 2.0 m, compute the wave energy. (Type in J, scientific notation OK — e.g. 1.16e-6)
Question 4. If the amplitude is doubled (everything else fixed), by what factor does the wave energy change? (Type the factor — e.g. 4)
Question 5. If the frequency is doubled (everything else fixed), by what factor does the wave energy change? (Type the factor)
Question 6. The wavelength is the distance between two consecutive crests, or how many times the distance between adjacent nodes? (Type the number)
Question 7 — Challenge. A wave transfers energy through a medium. Does it also transport the matter of the medium along with it? (Answer yes or no, with the key idea)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission, and include labelled screenshots of the standing wave and ruler for each scenario.

Wavelength and Energy of a Wave

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To measure the wavelength of a standing wave on a string with the ruler, compare it with the theoretical value λ = 2L/n, and calculate the energy the wave carries using K = ¼μA²ω²λ for two scenarios with different amplitude and frequency.

Theory

A standing wave on a string of length L has wavelength λ = 2L/n, where n is the number of segments. The angular frequency is ω = 2πf, and the energy of the wave is K = ¼μA²ω²λ, with μ the linear mass density. The energy is proportional to the square of the amplitude and the square of the frequency.

λ = 2L/n · ω = 2πf · K = ¼μA²ω²λ
μ = 0.0015 kg/m, L = 2.0 m

Results

ScenarioA (m)f (Hz)ω (rad/s)λₑ (m)λₜ (m)μ (kg/m)Kₑ (J)Kₜ (J)
10.00501.257.852.002.000.00151.16×10⁻⁶1.16×10⁻⁶
20.00752.2514.142.002.000.00158.43×10⁻⁶8.43×10⁻⁶

Sample calc, Scenario 1: ω = 2π(1.25) = 7.85 rad/s; K = ¼(0.0015)(0.0050)²(7.85)²(2.00) = 1.16×10⁻⁶ J.

Discussion

The measured wavelengths matched the theoretical λ = 2L/n closely (2.00 m for n = 2 in both scenarios), with any small differences coming from reading the ruler and judging the exact crest positions on screen. From Scenario 1 to Scenario 2 the amplitude rose by a factor of 1.5 and the frequency by 1.8; the energy increased from 1.16×10⁻⁶ J to 8.43×10⁻⁶ J — a factor of about 7.3, which is close to (1.5)²×(1.8)² = 7.3, confirming that the energy grows with the square of both amplitude and frequency. Higher tension produced cleaner, more stable standing waves, while increasing damping reduced the amplitude over time and made the pattern harder to read.

Conclusion

The experimental and theoretical wavelengths agreed, validating λ = 2L/n, and the calculated wave energy rose with the square of amplitude and frequency, consistent with K = ¼μA²ω²λ. Small differences arose from simulation resolution and human measurement error.

Practice Questions

Show all work and include units. Use λ = 2L/n, ω = 2πf, K = ¼μA²ω²λ, μ = 0.0015 kg/m unless told otherwise.

Question 1
A 3.0 m string vibrates in 3 segments. Find the wavelength and, if the wave speed is 12 m/s, the frequency.
Hint: λ = 2L/n, then f = v/λ.
Question 2
A wave has A = 0.0075 m, f = 2.25 Hz, λ = 2.0 m on a string with μ = 0.0015 kg/m. Calculate ω and the wave energy K.
Hint: ω = 2πf; K = ¼μA²ω²λ.
Question 3
Two waves are identical except one has triple the amplitude of the other. What is the ratio of their energies?
Hint: K ∝ A², so the ratio is 3² = 9.
Question 4
Explain why the energy of a wave depends on the square of the frequency rather than the frequency itself.
Hint: K contains ω² = (2πf)², so doubling f quadruples the energy.
Question 5
A string fixed at both ends shows 4 segments over a 2.0 m length. Find the wavelength and the distance between adjacent nodes.
Hint: λ = 2L/n; node spacing = λ/2.
Question 6 — Challenge
A wave's amplitude is halved while its frequency is doubled. Does the energy increase, decrease, or stay the same? By what overall factor?
Hint: A → ×¼ (½²), f → ×4 (2²); overall ¼ × 4 = 1, unchanged.