Lesson 3 of 5
Determining wave speed
I can define and use speed, frequency, wavelength, period and amplitude, including graphical representations.Syllabus
Syllabus K326 / K327, 9(c). Define and use speed, frequency, wavelength, period and amplitude, including graphical representations.
I can apply wave speed = frequency x wavelength in new situations.Syllabus
Syllabus K326 / K327, 9(d). Apply wave speed = frequency x wavelength in new situations.
Make a guess. You are not marked.
A signal generator drives sound into the same air at the same speed. Its frequency is doubled from 500 Hz to 1000 Hz. What happens to the wavelength?
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It halves, because the wavelength is the speed divided by the frequency.
v = f x wavelength and the speed is fixed by the air, so wavelength = v / f. Doubling f halves the wavelength: 0.68 m becomes 0.34 m at 340 m/s.
In one period, a repeating wave pattern advances by one wavelength. Its speed is wavelength divided by period.
Speed is distance divided by time. For a periodic wave, the distance is one wavelength λ and the time is one period T. Frequency f is 1/T.
Worked example
Use the rope graph values
The period is 0.40 s and the wavelength is 0.80 m.
- Frequency: f = 1/0.40 = 2.5 Hz.
- Wave speed: v = 2.5 x 0.80 = 2.0 m/s.
- Check the meaning: in 0.40 s the pattern advances 0.80 m, giving the same speed 0.80/0.40 = 2.0 m/s.
Try it step by step
Work it out in steps: Frequency f = 1 / T in Hz, then Wave speed v = f x wavelength in m/s.
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Problem 1.
Frequency f = 1 / T in Hz: f = 1 / 0.20 = 5.0 Hz
Wave speed v = f x wavelength in m/s: v = 5.0 x 1.5 = 7.5 m/s (2 sf)
Problem 2.
Frequency f = 1 / T in Hz: f = 1 / 0.050 = 20 Hz
Wave speed v = f x wavelength in m/s: v = 20 x 0.40 = 8.0 m/s (2 sf)
Problem 3.
Frequency f = 1 / T in Hz: f = 1 / 0.025 = 40 Hz
Wave speed v = f x wavelength in m/s: v = 40 x 0.30 = 12 m/s (2 sf)
Rearrange for the unknown
A wave with speed 3.6 m/s and frequency 12 Hz has wavelength 3.6/12 = 0.30 m. Dividing frequency by speed would not give a length.
Controlled comparison
Higher frequency at unchanged sound speed
Use a supplied sound speed of 340 m/s under unchanged propagation conditions.
- At 500 Hz, λ = 340/500 = 0.68 m.
- At 1000 Hz, λ = 340/1000 = 0.34 m.
Doubling frequency halves wavelength because the speed is held fixed. Do not apply that conclusion without checking the stated speed or conditions; not every possible water-wave comparison has frequency-independent speed.
Measure several wavelengths and several cycles
For a steady ripple-tank pattern at fixed water depth, measure the separation of equivalent crests in a correctly calibrated top view. Time complete source oscillations separately. The following are supplied model readings.
Six crests enclose five wavelengths
A supplied calibrated span is 40.0 cm from crest 1 to crest 6. Use that reading, not the displayed size of the picture.
One wavelength is 40.0/5 = 8.00 cm = 0.0800 m. Count the five gaps between the six crests rather than dividing by the number of crests.
| Trial | Time / s |
|---|---|
| 1 | 4.90 |
| 2 | 5.00 |
| 3 | 5.10 |
The mean time is (4.90 + 5.00 + 5.10)/3 = 5.00 s. Start and finish the count at the same stage of the source's motion.
T = 5.00/20 = 0.250 s
v = 4.00 x 0.0800 = 0.320 m/sThe 5.00 s interval is for 20 oscillations. It is not the time for one crest to travel from the first to the sixth position.
Choose improvements that address a cause
- Uncertain crest positions: measure a longer span containing several wavelengths and divide by the number of intervals. A similar endpoint-reading uncertainty then forms a smaller fraction of the full span.
- Start/stop reaction time: time several complete cycles. This reduces the fractional effect of reaction time compared with timing only one cycle.
- Incorrect length scale: calibrate the image or ruler in the observed plane and read without parallax. A projected image need not have the same physical scale as the water surface.
- Unclear pattern: keep the source steady and the water depth unchanged, and avoid an area where reflected waves overlap the incident pattern.
Repetition shows the variation in readings. It does not correct counting six intervals instead of five, using an incorrect scale, or changing the propagation conditions between the length and time measurements.
A different direct method follows the same pulse between two marked positions and uses distance divided by travel time. That travel time is not automatically the oscillation period. Identify which event starts and stops the timing.
Keep a spacing measurement separate from a cycle count. First find one wavelength and the number of cycles per second; then use v = fλ.
Watch out for this
A student says: "Doubling the frequency of a sound doubles its speed through the air." What is wrong with this?
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The speed of sound is set by the air, not by the source. With v = f x lambda and v unchanged, doubling the frequency halves the wavelength.
Check your understanding
The span from the first to the sixth crest is 40.0 cm. The source completes 20 oscillations in a mean time of 5.00 s. What is the wave speed for these readings?
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0.320 m/s, using five wavelengths and frequency 4.00 Hz.
Wavelength = 0.400/5 = 0.0800 m. Frequency = 20/5.00 = 4.00 Hz. Therefore v = f lambda = 4.00 x 0.0800 = 0.320 m/s. The 5.00 s interval times source oscillations, not one crest travelling across the measured span.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
Water waves travel from deep water into shallower water near a beach. State and explain what happens to their speed, frequency and wavelength. [3]
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Exam-style question
In a ripple tank, 10 complete waves pass a point in 4.0 s. The distance between neighbouring crests is 1.6 cm. Calculate the frequency and the speed of the waves. [3]
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Exam-style question
Data-based question: Waves in a ripple tank [9]
A student uses a vibrating dipper to make straight water waves in a ripple tank of uniform depth. She changes the frequency of the dipper and measures the wavelength each time.
| frequency / Hz | wavelength / cm |
|---|---|
| 5.0 | 6.0 |
| 10 | 3.0 |
| 15 | 2.0 |
| 20 | 1.5 |
- (a) Calculate the wave speed for each frequency, in m/s. [2]
- (b) Describe the relationship between frequency and wavelength shown by the data. [1]
- (c) Predict the wavelength when the frequency is 25 Hz. [1]
- (d) The water is made shallower. At 10 Hz the wavelength is now 2.4 cm. Calculate the new wave speed and state what this shows. [2]
- (e) A student says: "The waves travel faster when the frequency is higher, because the dipper vibrates faster." Use the data to evaluate this claim. [2]
- (f) State one assumption you made in (c). [1]
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