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Questions on progressive waves
A progressive wave transfers energy through a medium while each particle oscillates about a fixed equilibrium position and does not travel with the wave. Amplitude, wavelength, frequency, period and phase difference describe it, linked by c = fλ and f = 1/T. Wavelength is read off a displacement-distance graph, period off a displacement-time graph or an oscilloscope trace.
21 original questions · 59 marks · the progressive waves notes · Waves
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State what a progressive wave transfers, and what the particles of the medium do.
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A progressive wave transfers energy (and information) through the medium (1). The particles oscillate about fixed equilibrium positions; they do not travel with the wave (1).A wave has frequency 50 Hz and wavelength 6.0 m. Calculate its speed.
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c = fλ = 50 × 6.0 (1)
c = 300 m s−1 (1)A wave has a period of 0.020 s. Calculate its frequency.
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f = 1/T = 1/0.020 (1)
f = 50 Hz (1)Define the amplitude of a progressive wave.
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The maximum displacement of a particle of the medium from its equilibrium (rest) position (1).The displacement–time graph for one point on a rope repeats every 0.25 s. State the name given to this repeat time and calculate the frequency of the wave.
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The repeat time is the period of the wave (1). f = 1/T = 1/0.25 = 4.0 Hz (1).State, in radians, the phase difference between two points on a progressive wave that are half a wavelength apart.
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π rad (the points are in antiphase) (1).A sound wave travels at 340 m s−1 with a frequency of 170 Hz. Calculate its wavelength.
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λ = c/f = 340/170 (1)
λ = 2.0 m (1)Red light of wavelength 600 nm travels at 3.0 × 108 m s−1. Calculate its frequency.
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f = c/λ = (3.0 × 108)/(600 × 10−9) (1)
f = 5.0 × 1014 Hz (1)A wave has a frequency of 25 Hz and a wavelength of 1.2 m. Calculate (a) its speed and (b) its period.
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(a) c = fλ = 25 × 1.2 (1)
c = 30 m s−1 (1)
(b) T = 1/f = 1/25 = 0.040 s (1)Two points on a progressive wave are 0.75 m apart. The wavelength is 1.0 m. Calculate the phase difference between the two points, in degrees.
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Phase difference = (Δx/λ) × 360° (1)
= (0.75/1.0) × 360° (1)
= 270° (1)A dolphin emits ultrasound of frequency 120 kHz. The ultrasound travels through seawater at 1500 m s−1. Calculate its wavelength.
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λ = c/f = 1500/(120 × 103) (1) = 0.0125 m (= 12.5 mm) (1).A displacement–distance graph of a wave on a wire shows a vertical distance of 24 mm from a trough to the next crest, and a distance of 0.80 m between adjacent crests. The frequency of the wave is 5.0 Hz. Determine the amplitude of the wave and calculate its speed.
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Amplitude = half the trough-to-crest distance = 24/2 = 12 mm (1). Wavelength = crest-to-crest distance = 0.80 m, so c = fλ = 5.0 × 0.80 (1) = 4.0 m s−1 (1).The transducer of an ultrasonic cleaning bath oscillates with a period of 25 μs. The sound it produces travels through the cleaning fluid at 1480 m s−1. Calculate the frequency of the sound and its wavelength in the fluid.
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f = 1/T = 1/(25 × 10−6) = 4.0 × 104 Hz (40 kHz) (1). λ = c/f = 1480/(4.0 × 104) (1) = 0.037 m (1).A signal is displayed on an oscilloscope. One complete cycle of the trace spans 5.0 divisions of the screen, and the time-base is set to 2.0 ms per division. Calculate the period and the frequency of the signal.
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T = divisions per cycle × time-base setting = 5.0 × 2.0 ms (1) = 10 ms = 0.010 s (1). f = 1/T = 1/0.010 = 100 Hz (1).On a lake, 20 complete waves pass a fixed post in 10 s, and successive crests are 3.0 m apart. Calculate (a) the frequency, (b) the wave speed and (c) the period.
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(a) f = 20/10 = 2.0 Hz (1)
(b) c = fλ = 2.0 × 3.0 (1)
c = 6.0 m s−1 (1)
(c) T = 1/f = 0.50 s (1)Two points on a wave of wavelength 2.4 m have a phase difference of π/2 radians. Calculate the distance between the points. Determine the phase difference, in degrees, between two points on the same wave that are 3.0 m apart.
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π/2 rad is a quarter of a cycle (1)
separation = λ/4 = 2.4/4 = 0.60 m (1)
For 3.0 m: Δx/λ = 3.0/2.4 = 1.25 cycles (1)
phase difference = 1.25 × 360° = 450°, equivalent to 90° (1)Explain why the particles in a progressive wave do not travel along with the wave, even though the wave itself moves forward.
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Each particle oscillates about its own fixed equilibrium position (1), passing energy to the next particle (1). It is this disturbance (energy) that moves forward through the medium; the particles themselves only move back and forth and stay in the same average place (1).Points P and Q are 1.8 m apart on a progressive wave of wavelength 2.4 m. A student claims that P and Q oscillate in antiphase. Deduce whether the student is correct.
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Phase difference = (Δx/λ) × 360° = (1.8/2.4) × 360° (1) = 270° (3π/2 rad) (1). Antiphase requires 180° (π rad), so the student is not correct: P is three-quarters of a cycle out of phase with Q (1).A ship's radar transmits electromagnetic waves of frequency 9.4 GHz. Electromagnetic waves travel at 3.0 × 108 m s−1. Show that the wavelength of the radar waves is about 3 cm. Go on to calculate the period of the waves.
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λ = c/f = (3.0 × 108)/(9.4 × 109) (1) = 0.032 m = 3.2 cm, which is about 3 cm (1). T = 1/f = 1/(9.4 × 109) (1) = 1.1 × 10−10 s (1).A machine generates continuous waves on a long stretched rope. A student is given two graphs: a displacement–distance graph showing the whole rope at one instant, and a displacement–time graph for a single point on the rope. Explain how the student can use the two graphs together to determine the speed of the waves, and explain why neither graph on its own is sufficient.
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The displacement–distance graph is a snapshot of every point at one instant, so its repeat distance gives the wavelength λ (1). The displacement–time graph follows one point, so its repeat time gives the period T (1). The frequency follows from f = 1/T (1) and the speed from c = fλ (or c = λ/T) (1). The distance graph carries no time information, so it cannot give T or f (1). The time graph carries no distance information, so it cannot give λ; only the two graphs together supply both quantities needed for the speed (1).A sinusoidal trace on an oscilloscope stands 6.0 divisions from its lowest point to its highest, and two complete cycles span 8.0 divisions. The y-gain is set to 0.50 V per division and the time-base to 5.0 ms per division. Calculate the amplitude and the frequency of the signal.
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The 6.0 divisions is the peak-to-peak height, and the amplitude is measured from the centre line to a peak, so it is 3.0 divisions (1); amplitude = 3.0 × 0.50 = 1.5 V (1). One cycle spans 8.0/2 = 4.0 divisions, so T = 4.0 × 5.0 = 20 ms = 0.020 s (1); f = 1/T = 50 Hz (1).
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