Physics • Year 11 • Module 3 • Lesson 2
Wave Properties and the Wave Equation
Build HSC Band 5–6 extended-response technique on interpreting wave graphs, applying the wave equation across boundaries, and evaluating phase relationships in scientific contexts.
1. Data + scenario: AM versus FM radio waves (Band 5–6)
8 marks Band 5–6
Scenario. Australian radio broadcasters use two main bands: AM (Amplitude Modulation) and FM (Frequency Modulation). ABC Radio National AM broadcasts at approximately 576 kHz; Triple J FM broadcasts at approximately 107 MHz. Both signals travel at the speed of light in air: v = 3.0×108 m/s. The table below summarises key data.
| Property | AM (ABC Radio National) | FM (Triple J) |
|---|---|---|
| Frequency | 576 kHz = 5.76×105 Hz | 107 MHz = 1.07×108 Hz |
| Wave speed in air | 3.0×108 m/s | 3.0×108 m/s |
| Wavelength (λ) | calculate | calculate |
| Period (T) | calculate | calculate |
Illustrative values based on Australian radio band allocations.
Q1. Analyse and evaluate the wave properties of AM and FM radio using the data above. In your response you must:
- Calculate the wavelength and period of both signals, showing full working.
- Use the wave equation to explain why AM signals have a much longer wavelength than FM signals despite travelling at the same speed.
- The lesson states that longer wavelengths diffract more easily around hills. Using wave properties (not electronics), explain why AM can reach rural communities that FM cannot.
- A student claims that the AM signal travels faster than the FM signal because it has a longer wavelength. Evaluate this claim using the wave equation.
- State one limitation of using wavelength alone to explain the difference in reception range.
2. Experimental design — verifying the wave equation using a ripple tank (Band 5–6)
7 marks Band 5–6
Research question. A Year 11 class wants to verify that v = fλ holds for water waves in a ripple tank by independently measuring wave speed, frequency and wavelength, then comparing the product fλ to the measured speed.
Constraints: You have access to a ripple tank with a motorised vibrating bar, a strobe light, a ruler, a stopwatch, a camera or phone camera for top-down video, and a light source below the tank to project shadows on a screen.
Q2. Design the investigation and present it in the format below.
- State a hypothesis (a testable prediction including what you expect to find if v = fλ holds).
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you would measure wave speed, frequency and wavelength independently.
- Explain what result would falsify the hypothesis.
- State two sources of uncertainty in your design and one way to reduce each.
Q1 — Sample Band 6 response (8 marks), annotated
Calculations:
AM: λ = v/f = 3.0×108 / 5.76×105 ≈ 521 m; T = 1/f = 1/(5.76×105) ≈ 1.7×10−6 s. [1 mark for both AM values correct with working]
FM: λ = 3.0×108 / 1.07×108 ≈ 2.8 m; T = 1/(1.07×108) ≈ 9.3×10−9 s. [1 mark for both FM values correct with working]
Why AM has longer wavelength: Both signals travel at v = 3.0×108 m/s in air — speed is determined by the medium (air/vacuum), not the source. From λ = v/f, at constant speed, frequency and wavelength are inversely proportional. AM has roughly 185 times lower frequency than FM, so it has roughly 185 times longer wavelength (521 m vs 2.8 m). [1 mark]
Diffraction and rural reception: Wave diffraction increases when wavelength is comparable to or larger than the obstacle size. Hills in rural Australia are typically hundreds of metres high and wide. An AM wavelength of ~520 m is comparable to or larger than many hills, so AM waves diffract around them and reach communities beyond the direct line of sight. FM wavelengths (~2.8 m) are far smaller than hills, so they diffract very little around them and require line-of-sight reception. [1 mark]
Evaluating the speed claim: The claim is incorrect. Both AM and FM are electromagnetic waves travelling in air; their speed is 3.0×108 m/s regardless of frequency. The wave equation v = fλ shows that if both waves travel at the same speed, a longer wavelength simply means lower frequency, not higher speed. Wavelength and frequency are inversely proportional at constant speed. [1 mark]
Limitation: Wavelength alone cannot fully explain reception range. Other factors include transmitter power (AM transmitters typically broadcast at much higher power than FM), ionospheric reflection of AM signals over long distances, and terrain-specific diffraction effects that vary with local geography. [1 mark]
Marking criteria (8 marks): 1 = AM λ and T calculated correctly; 1 = FM λ and T calculated correctly; 1 = explains inverse f–λ relationship at constant speed; 1 = diffraction argument linked to wavelength-vs-obstacle size; 1 = correctly refutes speed claim using v = fλ and fixed medium speed; 1 = valid limitation stated clearly; 1 = response uses precise terminology throughout (frequency, wavelength, wave speed, diffraction, medium, inversely proportional); 1 = response integrates all five required criteria logically.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If v = fλ holds for water waves, then the product of independently measured frequency (f) and wavelength (λ) will equal the independently measured wave speed (v) within experimental uncertainty. IV = vibration frequency of the bar (set by motor speed). DV = ratio of measured speed to calculated f×λ. Controlled variables: water depth (affects wave speed), amplitude of vibration (amplitude should not affect f or λ), temperature of water. [1 mark — hypothesis with IV, DV, two controlled variables]
Procedure: (1) Set the ripple tank to a fixed water depth (e.g. 5 mm); set the vibrator to a frequency of approximately 10 Hz and note the motor setting. (2) Measure wavelength: use the projected shadow pattern on the screen to identify five crests; measure the total crest-to-crest distance across five wavelengths with a ruler, then divide by 5 to reduce parallax error. (3) Measure frequency: adjust the strobe light frequency until the wave pattern appears stationary — the strobe frequency equals the wave frequency. Cross-check by counting crests passing a fixed reference dot in the tank over 10 seconds using slow-motion video. (4) Measure wave speed: mark two points 50 cm apart on the tank floor; use the slow-motion video to time a crest travelling between the marks; v = distance/time. Repeat three times and average. (5) Calculate f×λ and compare to the measured v. [1 mark — four steps including independent speed and wavelength measurement]
Falsification: If f×λ consistently differs from the directly measured v by more than the experimental uncertainty (i.e. the discrepancy is not explained by measurement errors), the hypothesis that v = fλ would be falsified. [1 mark]
Uncertainty 1: Parallax error when measuring wavelength from the projected shadow, because the camera/eye position relative to the screen affects apparent crest position. Reduction: position the camera directly above (top-down) and use a grid scale drawn on the screen to calibrate distance. [1 mark]
Uncertainty 2: Timing the crest movement for wave speed is affected by reaction time with a stopwatch, especially for fast waves. Reduction: use slow-motion video (240 fps) and count frames between a crest passing each marked point, then convert frames to seconds using the known frame rate. [1 mark]
What results would show: If v = fλ, then increasing the motor frequency (higher f) while keeping water depth (and therefore wave speed) constant should produce shorter wavelength, with f×λ remaining equal to the measured speed. [1 mark]
Marking criteria (7 marks): 1 = testable hypothesis naming IV and DV; 1 = four steps with independent measurement of v, f and λ described; 1 = states what would falsify the hypothesis; 1 = one valid uncertainty with a specific reduction strategy; 1 = second valid uncertainty with a specific reduction strategy; 1 = explains expected result consistent with v = fλ; 1 = precise physics terminology throughout.