Flashcards · 9702 · AS Topic 7

Waves (AS Level) flashcards

25 cards on AS Topic 7, subtopics 7.1 to 7.5. Pick a subtopic or study them all.

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Define amplitude, wavelength, period and frequency: Amplitude: maximum displacement from equilibrium. Wavelength λ: distance between adjacent points in phase. Period T: time for one oscillation. Frequency f: oscillations per second (Hz).
Phase difference vs path difference: Phase difference: how far out of step two points or waves are, as an angle (360° or 2π rad per cycle). Path difference: a distance. They are related through the wavelength.
Derive v = fλ: In one period T the wave travels one wavelength λ, so v = λ ÷ T. Since f = 1 ÷ T, v = fλ.
Intensity of a wave, and how it depends on amplitude: Intensity = power ÷ area (W m⁻²). Intensity ∝ amplitude², so doubling the amplitude quadruples the intensity.
CRO: peak 2.5 divisions high at 2.0 V/div; one cycle 4.0 divisions at 5.0 ms/div. Find amplitude and frequency.: Amplitude = 2.5 × 2.0 = 5.0 V. T = 4.0 × 5.0 ms = 0.020 s, so f = 1 ÷ T = 50 Hz.
What does a progressive wave transfer?: Energy, from source to receiver, without transferring matter: each part of the medium oscillates about a fixed position.

Transverse wave: The oscillations are perpendicular to the direction of energy transfer. Crests and troughs. Examples: all electromagnetic waves, waves on a rope, water waves.
Longitudinal wave: The oscillations are parallel to the direction of energy transfer. Compressions and rarefactions. Examples: sound, ultrasound.
How can a slinky spring show both types of wave?: Pushed and pulled along its length: longitudinal (compressions and rarefactions). Moved side to side: transverse.
How is a ripple tank used to measure the speed of water waves?: A stroboscope flashing at the wave frequency "freezes" the pattern so λ can be measured (across several waves, then divided). Then v = fλ.
Which type of wave can be polarised?: Only transverse waves. A longitudinal wave oscillates only along its direction of travel, so there is no plane to select.

Doppler equation for a moving source and a stationary observer: f₀ = fₛv ÷ (v ± vₛ), where v is the wave speed and vₛ the source speed. Source approaching: minus (higher observed frequency). Source receding: plus (lower observed frequency).
Explain why an approaching siren sounds higher pitched: The wavefronts ahead of the moving source are bunched together, so the observed wavelength is shorter and the observed frequency higher. The source frequency itself is unchanged.
A 500 Hz horn approaches at 20 m s⁻¹ (sound speed 340 m s⁻¹). Find the observed frequency.: f₀ = 500 × 340 ÷ (340 − 20) = 500 × 340 ÷ 320 = 531 Hz.
The same 500 Hz horn moves away at 20 m s⁻¹ (sound speed 340 m s⁻¹). Find the observed frequency.: f₀ = 500 × 340 ÷ (340 + 20) = 500 × 340 ÷ 360 = 472 Hz.

The electromagnetic spectrum in order of decreasing wavelength: Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. Frequency increases in the same order.
Properties shared by all electromagnetic waves: They are transverse and travel at the same speed in a vacuum, c = 3.00 × 10⁸ m s⁻¹.
Wavelength range of visible light: 400 nm (violet) to 700 nm (red).
Approximate wavelengths of microwaves, infrared and ultraviolet: Microwaves: about 1 mm to 0.1 m. Infrared: about 700 nm to 1 mm. Ultraviolet: about 10 nm to 400 nm. (The boundaries are not sharp.)
Approximate wavelengths of radio waves, X-rays and gamma rays: Radio: longer than about 0.1 m. X-rays: about 10 pm to 10 nm. Gamma rays: shorter than about 10 pm. (The regions overlap at their boundaries.)

Polarisation: Restricting the oscillations of a transverse wave to a single plane (containing the direction of travel).
Malus’s law: I = I₀ cos²θ, for plane-polarised light of intensity I₀ passing through a polarising filter whose transmission axis is at θ to the plane of polarisation.
Plane-polarised light of 12 W m⁻² meets a filter at 60° to its plane of polarisation. Find the transmitted intensity.: I = 12 × cos²60° = 12 × 0.25 = 3.0 W m⁻².
What intensity passes through two crossed polarisers (at 90°)?: Zero: cos²90° = 0 exactly.
Why can light be polarised but sound cannot?: Light is transverse, so it can oscillate in many planes and one can be selected. Sound is longitudinal: its oscillation is along the direction of travel, so there is no plane to filter.