Physics homework
Wave speed homework: distance or time on the axis?
Choose the right wave relationship, count complete cycles, convert units and check your own working before asking for an AI hint.
Wave speed homework often goes wrong before the arithmetic starts. A curve may show displacement against distance, or displacement against time. Those pictures can look alike while giving different quantities. Reading a gap as metres when the axis is seconds produces an answer that no calculator can repair. Start by identifying what is measured, then choose the relationship that matches the information actually supplied.
This guide uses original, fictional classroom data and free practice questions you can work on paper without an account, app or paid worksheet. The photographs illustrate study habits; they are not calibrated wave diagrams or evidence from an experiment. Lirno can help with a focused hint, explanation or check where your school permits it. Keep the original task and your own attempt visible so that any help addresses your particular uncertainty.
1. Start wave speed homework with three different quantities
Wave speed v describes how quickly a repeated disturbance travels through the medium. Frequency f counts complete oscillations per second at a fixed position. Wavelength λ, the Greek letter lambda, is a spatial repeat distance along the direction of travel. A point on a rope can oscillate up and down while the disturbance moves along the rope. Its instantaneous vertical motion is not the propagation speed you are calculating.
For a periodic travelling wave, use v = f × λ. If v is in metres per second and f in hertz, λ must be in metres. One hertz means one cycle per second, so the units multiply to metres per second. In a simple nondispersive school model, keeping the same medium and conditions fixes the speed; changing frequency then changes wavelength inversely. Do not assume that every real medium is nondispersive.
The OpenStax explanation of wave quantities defines the spatial and temporal repeats and connects them through propagation speed. Your worksheet may write c or u instead of v. Follow its definitions: the letter c does not automatically mean the vacuum speed of light in a sound problem. Underline the requested quantity and list the given values with their symbols and units before substituting anything.
2. Keep the whole task in view when you scan
A useful reading includes the written question, graph title, horizontal-axis label, scale marks, units and any note about the medium. Copy these details beside your working. A number called amplitude is a maximum displacement, not a wavelength. A stated travel distance is not necessarily one full cycle either. If a caption says that a drawing is schematic, do not turn its apparent centimetres on your screen into physical measurements.
If permitted, use Lirno’s homework scanning workflow with the page flat, evenly lit and fully framed. Include small prefixes such as k in kHz and m in ms, as well as any arrows identifying a measured span. Remove names or unrelated private information first. A close crop of the curve alone may remove the one axis label that determines the whole method.
Compare the recognised task with the original. Check λ against a misread letter, 0.80 against 80, and milliseconds against metres. Then ask a narrow question: “Does this horizontal axis give distance or time, and which quantity can I read from it?” Correct a recognition error before accepting an explanation. A plausible sentence about waves cannot compensate for a missing axis or an invented scale.
3. Rearrange the relationship before entering the numbers
For an original exercise, suppose a periodic disturbance travels at 24 m/s and has frequency 6.0 Hz. The wavelength is λ = v ÷ f = 24 ÷ 6.0 = 4.0 m. Write the symbolic rearrangement first. Multiplying 24 by 6.0 would give 144 numerically, but the units would be metres per second squared, which is not a wavelength. Units help expose the wrong operation.
If the same task instead supplies λ = 4.0 m and f = 6.0 Hz, calculate v = 6.0 × 4.0 = 24 m/s. If it supplies v and λ, use f = v ÷ λ to recover 6.0 Hz. These are rearrangements of one relationship, not three unrelated rules to memorise. A reverse substitution should reproduce the given quantity without changing the original values.
Use the exact speed stated for that exercise. For a fictional sound question giving v = 320 m/s and f = 800 Hz, λ = 0.40 m. This does not claim that sound always travels at 320 m/s: its speed depends on the medium and conditions. Similarly, use a vacuum light speed only when the problem identifies electromagnetic propagation in vacuum or explicitly supplies that approximation.
4. Decide whether the graph gives wavelength or period
On a displacement–distance snapshot, adjacent crests have the same phase and are separated by one wavelength. Adjacent troughs work too. A crest to the next trough is only half a wavelength for a sinusoidal wave. In an original snapshot, the first and fifth consecutive crests are 1.6 m apart. There are four complete repeat intervals, so λ = 1.6 ÷ 4 = 0.40 m, not 0.32 m.
On a displacement–time graph at one fixed point, the corresponding repeat interval is the period T. Suppose the first and sixth consecutive crests are separated by 0.025 s. Five periods fit into that span: T = 0.025 ÷ 5 = 0.0050 s. Hence f = 1 ÷ T = 200 Hz. Count spaces between matching points rather than counting the points themselves.
If those two graphs describe the same periodic wave under the same conditions, combine λ = 0.40 m and f = 200 Hz to obtain v = 80 m/s. Without that connection, do not combine numbers from unrelated panels. A spatial snapshot alone cannot determine propagation speed without time or frequency information. A time trace alone cannot supply wavelength without speed or suitable spatial information.
5. Convert prefixes once and preserve the units
In the sound example, a frequency of 0.80 kHz equals 800 Hz. Using the supplied 320 m/s gives λ = 320 ÷ 800 = 0.40 m. Entering 0.80 without converting would produce 400 m, a factor of a thousand too large. Write the converted value on its own line and use that line consistently; do not convert once in your notes and a second time in the calculator.
A period of 5.0 ms equals 0.0050 s, so f = 200 Hz. Taking the reciprocal of the unconverted number gives 0.20 per millisecond, which equals 200 per second, not 0.20 Hz. You can work in other compatible units, but keeping metres, seconds and hertz makes a school solution easier to audit. State the final length unit even when the calculator displays a tidy decimal.
For a separate vacuum radio example, let v = 3.0 × 10⁸ m/s and f = 75 MHz = 7.5 × 10⁷ Hz. Then λ = 4.0 m. Keep scientific-notation powers inside the division and use parentheses when entering the denominator. Retain extra calculator digits until the end, then round according to the task and the precision of its data rather than claiming many unsupported decimal places.
6. Check what changes when the conditions change
For a simple nondispersive medium with fixed conditions and v = 12 m/s, a source frequency of 4.0 Hz gives λ = 3.0 m. Doubling frequency to 8.0 Hz gives λ = 1.5 m while the speed stays 12 m/s. The product fλ remains constant. This statement needs its fixed-speed assumption; it is not a universal rule that frequency can never affect propagation in more complicated wave systems.
Now consider a stationary boundary between two media, with a continuously driven source at 4.0 Hz. If the transmitted periodic wave travels at 8.0 m/s in the second medium, its wavelength is 2.0 m. In this ordinary boundary model the frequency stays tied to the source while speed and wavelength change. Do not reuse the first medium’s 3.0 m wavelength together with the second medium’s speed.
Amplitude answers a different question. A larger vertical displacement does not by itself tell you a larger wavelength or a faster wave. Also distinguish travel speed from the speed of an individual vibrating point. If a task moves on to standing waves, harmonics or Doppler shifts, identify the extra conditions first. The simple substitution here does not establish which fraction of a wavelength fits into a resonating length.
7. Ask for the first uncertain step, then verify it
Show Lirno Tutor your attempt with the axis interpretation, converted values, chosen relationship and final unit. A useful request is “I counted five crests across 1.6 m and divided by five; which interval have I counted incorrectly?” Another is “Check only my kHz conversion before I calculate wavelength.” Ask for a hint first, work the next line yourself, then request an explanation if the confusion remains.
Compare any reply with the page and your reverse calculation. AI can misread a graph, confuse period with frequency or reason incorrectly. Lirno does not guarantee correctness, grades, mastery or permission to use it. Follow your school’s rules and ask your teacher if a diagram or instruction remains ambiguous. A response that agrees with your answer still needs a sound explanation and consistent units.
Lirno is free to download or start; some AI usage levels or advanced features may require Premium. The paper questions in this guide require neither purchase nor app access. After a useful hint, close the response and reconstruct the corrected step independently. Keep a brief note of the actual error, such as “five peaks make four intervals,” rather than copying a whole answer you cannot explain.
8. Try these free practice questions before reading the checks
First, a wave has v = 18 m/s and f = 3.0 Hz: find its wavelength. Second, a wave has λ = 0.25 m and f = 48 Hz: find its speed. Third, a time trace spans 12 ms between its first and fourth consecutive crests: find period and frequency. Write knowns, unknown, relationship, conversion, substitution and unit for each question before comparing answers.
The checks are λ = 6.0 m; v = 12 m/s; and T = 12 ÷ 3 = 4.0 ms = 0.0040 s, giving f = 250 Hz. For the third question, four crests enclose three intervals. If your result differs, identify whether the mismatch began with the interval count, prefix conversion or reciprocal. Do not simply replace the final line and leave the faulty reasoning above it.
Finally, test a claim: “A wave with twice the amplitude must have twice the speed.” The given information does not justify that conclusion in the simple linear model. You need information about propagation conditions, not just vertical size. Finish your own homework with a sentence explaining what each graph provides and why the chosen relationship applies. That explanation makes your answer checkable even when the numbers change.
