Waves, light and sound, Year 11: wave properties and the wave equation

A transverse wave

HintRipples spreading across a pond are the standard example.

The whyIn this kind of wave each point moves up and down (or side to side) while the energy moves along. Ripples on water are AQA's named example, and every electromagnetic wave is of this kind too. In the other kind, a longitudinal wave, the vibrations are along the direction of travel.

longitudinal

HintThe air is pushed and pulled along the same line that the sound is heading.

The whyIn a sound wave the air particles vibrate to and fro along the line the wave travels, which is what longitudinal means. This makes regions where the air is squashed (compressions) and regions where it is spread out (rarefactions). AQA names sound in air as the example to know.

A rarefaction

HintIts opposite is a compression; this is the stretched-out part of a slinky spring that is being pushed back and forth.

The whyA longitudinal wave is a pattern of squashed regions (compressions) and stretched regions (rarefactions) moving along. In a sound wave the air pressure is slightly above normal in a compression and slightly below it here. The distance from one such region to the next is one wavelength.

It bobs up and down but is not carried to the bank, so the water does not travel along with the wave

HintAsk whether the toy ends up any nearer the edge after ten ripples have gone by.

The whyA floating object is a marker for the water it sits on. It rises and falls as each ripple passes but stays in the same place, which is the evidence AQA asks for that the wave travels and the water does not. What the wave carries from place to place is energy.

The air only vibrates back and forth about a fixed position; it does not travel away from the speaker with the sound

HintCompare what the flame would do in a steady draught made by a fan.

The whyIf sound were a flow of air out of the speaker, the flame would lean one way all the time, as it does in a draught. Instead it moves to and fro, showing that the air particles vibrate along the direction of the sound while staying, on average, where they were. The wave and its energy travel; the air does not.

The amplitude of the wave

HintOn a diagram it is measured from the middle line to the top of a crest.

The whyAmplitude is measured from the undisturbed (rest) position to the top of a crest, or equally from the rest position to the bottom of a trough. A bigger amplitude means the wave carries more energy: a louder sound or a higher water wave. It is a distance, so its unit is the metre.

4; 25

HintOne answer needs a halving, the other a subtraction.

The whyAmplitude is measured from the rest position, which lies halfway between crest and trough, so it is half of 8 cm. Wavelength is the distance from one point on a wave to the same point on the next wave: here crest to crest, 35 cm − 10 cm.

5

HintFrequency is how many go by in each single second.

The whyFrequency counts how many waves go past a point every second, so divide the count by the time: 20 ÷ 4. One hertz (Hz) means one wave per second.

The period of the wave

HintIt is measured in seconds, and its symbol is T.

The whyPeriod and frequency describe the same thing from two sides: frequency is waves per second, period is seconds per wave. That is why period = 1 ÷ frequency (T = 1 ÷ f), with T in seconds and f in hertz. AQA gives this equation on the Physics equation sheet.

0.004 s

HintThe equation sheet links these two quantities with a 'one divided by'.

The whyPeriod = 1 ÷ frequency, or T = 1 ÷ f, with T in seconds (s) and f in hertz (Hz). So T = 1 ÷ 250 = 0.004 s, which is 4 milliseconds. AQA gives this equation on the Physics equation sheet: pupils select and apply it and need not recall it.

wave speed = frequency × wavelength; v = f × λ

HintTwo of the quantities multiply to give the one measured in m/s.

The whyThe speed v is measured in m/s (metres per second), the frequency f in Hz (hertz) and the wavelength λ, the Greek letter lambda, in m (metres). Every kind of wave obeys it. AQA expects this equation to be recalled as well as applied.

3 m

HintChange the frequency into hertz first, then rearrange the wave equation.

The why100 MHz is 100 × 10⁶ Hz = 1.0 × 10⁸ Hz. Rearranging v = f × λ gives λ = v ÷ f = (3.0 × 10⁸) ÷ (1.0 × 10⁸) = 3 m. The speed of radio waves is given in the question because the Trilogy specification does not quote it.

20

HintThe speed is in metres every second, so the other length must be in metres too.

The whyConvert first: 1.5 cm = 0.015 m. Rearranging v = f × λ gives f = v ÷ λ = 0.30 ÷ 0.015 = 20 Hz. Twenty ripples leave the dipper every second, which is why they cannot be counted by eye.

halves

HintWith the speed fixed, the two other quantities in the wave equation must multiply to the same number.

The whyWave speed = frequency × wavelength. If the speed cannot change, frequency and wavelength are inversely proportional: twice as many waves each second means each wave is half as long.

330 m/s

HintKilo means a thousand; convert before you multiply.

The why1.1 kHz = 1100 Hz. Wave speed = frequency × wavelength = 1100 × 0.30 = 330 m/s. That matches the typical value AQA gives for the speed of sound in air.

Human reaction time, which is large compared with the short time being measured. Use a much greater distance (and repeat and average)

HintSound covers that gap in about a third of a second. How quick is a thumb?

The whyAt about 330 m/s the sound takes only about 0.3 s to cover 100 m, and a person's reaction time is typically 0.2 s to 0.9 s, so the error can be as big as the reading. A longer distance gives a longer time, so the same reaction error is a smaller fraction of it. The speed is then distance ÷ time.

One wavelength is too small to measure accurately; measuring ten makes the percentage error ten times smaller

HintThink what a 1 mm slip on the ruler does to a 2 cm reading and to a 20 cm reading.

The whyThe ruler's uncertainty is about the same whatever is measured, perhaps a millimetre. Spread over ten waves, the uncertainty in each wavelength is a tenth as big. The same trick is used for timing: count many waves, then divide.

2.4; 3; 7.2

HintFind the length of one wave, then how many pass each second, then combine the two.

The whyWavelength = 24 ÷ 10, frequency = 30 ÷ 10, and wave speed = frequency × wavelength = 3 × 2.4. Because the wavelength was left in centimetres, the speed comes out in centimetres per second; in metres it is 0.072 m/s. This is the method of the ripple-tank required practical.

Photograph the pattern (with a ruler in view), or light it with a stroboscope so that it appears to stand still

HintYou need the moving shadows on the screen to be frozen somehow.

The whyA lamp above the tank casts the ripples as bright and dark lines on a screen below. A photograph freezes them so that several wavelengths can be measured against a ruler. A stroboscope flashing at the same frequency as the ripples does the same job, because each flash catches the pattern in the same place.

40 m/s

HintFind the length of one loop, then of two.

The whyThree loops fill 1.2 m, so one loop is 0.4 m and one wavelength (two loops) is 0.8 m. Wave speed = frequency × wavelength = 50 × 0.8 = 40 m/s. This is the 'waves in a solid' half of the required practical: the frequency is read from the signal generator and the wavelength is measured with a ruler.

330

HintThe equation is the Year 7 one for anything that covers a known distance in a known time.

The whySpeed = distance ÷ time = 1.65 ÷ 0.005 = 330 m/s, the typical value AQA gives for sound in air. Using microphones and an electronic timer removes human reaction time, which spoils the stopwatch method over short distances.