Wave properties

Calculate wave speed and describe transverse and longitudinal waves.

Start here

The key idea

Waves transfer energy without transferring matter overall.

Equation to know

wave speed = frequency x wavelength

Wave Properties
amplitudewavelength

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

Transverse and longitudinal waves

Waves transfer energy and information without transferring matter.In a transverse wave the oscillations are at right angles to the direction of energy transfer, such as water waves and all electromagnetic waves.In a longitudinal wave the oscillations are parallel to the direction of energy transfer, producing compressions and rarefactions, such as sound waves.In both types the particles or fields oscillate about a fixed point and do not travel with the wave.

2

Key wave quantities

Amplitude is the maximum displacement from the rest position.Wavelength is the distance between two corresponding points on adjacent waves, such as crest to crest.Frequency is the number of complete waves passing a point each second, measured in hertz (Hz).The period T is the time for one complete wave, and T = 1 / f.

3

The wave equation

The wave speed is linked to frequency and wavelength by wave speed = frequency x wavelength (v = f x lambda), measured in m/s.For a given medium the wave speed is constant, so if frequency increases the wavelength must decrease.This equation applies to all waves, including sound and electromagnetic waves.

4

Reflection, refraction and measuring speed

When a wave meets a boundary it can be reflected, transmitted or absorbed.Refraction is the change in direction of a wave when it crosses a boundary and changes speed, for example light bending as it enters glass.The speed of sound in air can be measured by timing an echo or by recording sound at two microphones a known distance apart.The speed of water waves can be found using v = f x lambda from a ripple tank.

Key terms

Definitions to learn

Amplitude

The maximum displacement of a point on a wave from its rest position.

Wavelength

The distance between two corresponding points on adjacent waves.

Frequency

The number of complete waves passing a point per second, in Hz.

Period

The time taken for one complete wave to pass a point, T = 1 / f.

Refraction

The change in direction of a wave as it changes speed crossing a boundary.

Worked example

A wave has frequency 50 Hz and wavelength 0.80 m. Calculate its speed.

1

Use v = f lambda.

2

Substitute the values.

Final answer

40 m/s

Exam habit

Write v = fλ before every wave calculation. Identify which two quantities are given, rearrange and solve.Distinguish wavelength (distance) from period (time): T = 1/f.

Watch out

Do not confuse amplitude with wavelength.

Examiner tips

How to score full marks

  • 1When using v = f x lambda, make sure frequency is in Hz and wavelength in metres before substituting.
  • 2Remember waves transfer energy, not matter — state this clearly if asked.
  • 3Amplitude relates to the energy/loudness of a wave; frequency relates to pitpitch, not amplitude.
Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A sound wave travels at 340 m/s and has frequency 680 Hz. Find its wavelength.
Mark scheme
  1. 1.Rearrange wavelength = speed / frequency.
0.50 m
2Describe one difference between transverse and longitudinal waves.
Mark scheme
  1. 1.Compare oscillation direction with energy-transfer direction.
Transverse oscillations are perpendicular to energy transfer; longitudinal oscillations are parallel.
3Explain why the frequency stays constant when a wave crosses a boundary.
Mark scheme
  1. 1.Think about the source.
The source fixes the frequency, although speed and wavelength may change.
4State one similarity and one difference between transverse and longitudinal waves.[2 marks]
Mark scheme
  1. 1.Give a similarity.
  2. 2.Give a difference.
Similarity: both transfer energy without transferring matter (1). Difference: in transverse waves oscillations are at right angles to energy transfer, in longitudinal waves they are parallel to it (1).
5A wave has a frequency of 50 Hz and a wavelength of 4 m. Calculate the wave speed.[2 marks]
Mark scheme
  1. 1.Use v = f x lambda.
  2. 2.Substitute.
v = f x lambda = 50 × 4 (1) = 200 m/s (1).
6A sound wave has a period of 0.004 s. Calculate its frequency.[2 marks]
Mark scheme
  1. 1.Use f = 1 / T.
  2. 2.Substitute.
f = 1 / T = 1 / 0.004 (1) = 250 Hz (1).
7A radio wave travels at 3×1083 \times 10^{8} m/s and has a frequency of 1.5×1081.5 \times 10^{8} Hz. Calculate its wavelength.[3 marks]
Mark scheme
  1. 1.Rearrange v = f x lambda to lambda = v / f.
  2. 2.Substitute.
  3. 3.Evaluate.
lambda = v / f = (3×1083 \times 10^{8}) / (1.5×1081.5 \times 10^{8}) (1) = 2 (1) m (1).
8A student measures the speed of sound using two microphones 1.2 m apart. The sound reaches the second microphone 0.0035 s after the first. Calculate the speed of sound and explain why repeating the measurement improves the result.[4 marks]
Mark scheme
  1. 1.Use speed = distance / time.
  2. 2.Substitute.
  3. 3.Evaluate.
  4. 4.Explain repeats reduce random error.
speed = distance / time = 1.2 / 0.0035 (1) = 343 m/s (accept 340-345) (1). Repeating the measurement and taking a mean reduces the effect of random errors (1) and makes the result more reliable/precise (1).
9A water wave in a ripple tank has a wavelength of 0.030 m and a frequency of 12 Hz. Calculate its speed. The frequency is then doubled while the medium stays the same. State the new wavelength.[2 marks]
Mark scheme
  1. 1.v = f x lambda = 12 × 0.030.
  2. 2.Speed stays the same when medium stays the same.
  3. 3.New wavelength = v / new f.
v = 12 × 0.030 = 0.36 m/s (1); speed stays constant in the same medium, so new lambda = v / (2f) = 0.36 / 24 = 0.015 m (1)
10Describe an experiment to measure the speed of sound in air using a starting pispistol and an echo from a large wall. Include measurements taken, how speed is calculated, and one precaution.[3 marks]
Mark scheme
  1. 1.Stand known distance from wall; fire pispistol.
  2. 2.Time between firing and hearing echo.
  3. 3.Speed = (2 x distance) / time.
  4. 4.Precaution: use a long distance to reduce timing error.
Stand a measured distance d from a large flat wall and fire a starting pispistol (1); use a stopwatch to measure the time t from the sound of the pispistol to hearing the echo from the wall; the sound travels to the wall and back, so speed = 2d / t (1); to reduce the percentage error in the timing, use as large a distance as possible (at least 50 m); repeating and averaging also reduces random error (1)
11Explain what happens to the speed, frequency and wavelength of a sound wave when it travels from air into water, where sound travels faster.[3 marks]
Mark scheme
  1. 1.Speed increases in water.
  2. 2.Frequency stays constant (determined by the source).
  3. 3.From v = f x lambda, if v increases and f stays constant, lambda must increase.
The speed of the sound wave increases as it enters water, because the water molecules transmit the compression more quickly (1); the frequency remains constant because it is set by the source and does not change as the wave crosses a boundary (1); since v = f x lambda and f is unchanged, the increase in speed means the wavelength must also increase (1)
12A student observes water waves in a ripple tank and measures 6 complete waves in a distance of 0.18 m passing a point in 2.0 s. Calculate the wavelength, frequency and speed of the waves.[3 marks]
Mark scheme
  1. 1.Wavelength = total distance / number of wavelengths = 0.18 / 6.
  2. 2.Frequency = number of waves / time = 6 / 2.0.
  3. 3.Speed = f x lambda.
lambda = 0.18 / 6 = 0.030 m (1); f = 6 / 2.0 = 3.0 Hz (1); v = 3.0 × 0.030 = 0.090 m/s (1)
13Explain, using the wave model, what refraction is and why it causes a wave to change direction when it crosses a boundary at an angle. Describe what happens to the wavefronts.[4 marks]
Mark scheme
  1. 1.Refraction: wave changes speed at boundary.
  2. 2.Part of wavefront enters new medium first and slows/speeds.
  3. 3.This causes wavefront to bend.
  4. 4.Direction of travel (ray) bends toward or away from normal.
Refraction occurs when a wave crosses a boundary between two media in which it travels at different speeds (1); when a wavefront hits the boundary at an angle, one side of the wavefront enters the new medium and changes speed before the other side; this causes the wavefront to tilt (1); the direction of energy travel (the ray) is perpendicular to the wavefront, so as the wavefront tilts the ray bends — towards the normal if the wave slows down, away from the normal if it speeds up (1); the frequency stays constant throughout, but the wavelength changes in proportion to the speed change (1)
14Ultrasound pulses are used to measure the depth of the sea from a ship. A pulse is sent down and the echo is detected 0.30 s later. The speed of sound in seawater is 1500 m/s. Calculate the depth of the sea below the ship. Suggest one reason why this method might give an inaccurate depth reading.[3 marks]
Mark scheme
  1. 1.Total distance = speed x time.
  2. 2.Depth = total distance / 2 (pulse goes down and back).
  3. 3.Inaccuracy: temperature/salinity changes affect speed.
Total distance = 1500 × 0.30 = 450 m (1); depth = 450 / 2 = 225 m (1); the speed of sound in seawater varies with temperature and salinity — if the actual speed differs from the assumed 1500 m/s, the calculated depth will be inaccurate (1)
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