Objectives

  • To select and use the wave equation v = f\lambda

Deducing and using the wave equation

Justification/deduction of the wave equation  v = f \lambda

Lets use an example of the coaches of a train are going past; You count how many coaches go by in a second and you know the length of one – so you multiply the two together to get the train’s speed.
Apply this to waves: count the number of waves passing each second (= frequency), and multiply by the length of each (= wavelength) to find the speed.

Speed = distance / time

Speed = wavelength / (1 / frequency)

Speed = frequency \times wavelength

v = f \lambda

This equation becomes:

c = f \lambda

for the speed of light, c.

You need to be able to select, use and rearrange this equation to find either  c ,  f  or  \lambda .

Using the equation  v = f \lambda ;

The speed of a sound wave can be determined if the frequency and wavelength are known.

Example 1
A sound wave, travelling through air of frequency 1200 Hz and wavelength of  28 cm travels at what speed?
v = ?
f = 1200 Hz
\lambda = 28 cm = 0.28 m – Don’t forget to change the wavelength to metres!

v = f \lambda
v = 1200 \times 0.28
v = 336 m/s

Example 2
A different sound wave, travelling through air has a frequency of  3750 Hz and a wavelength of  9 cm . What speed does this wave travel at?
v = ?
f = 3750 Hz
\lambda = 9 cm = 0.09 m

v = f \lambda
v = 3750 \times 0.09
v = 337.5 m/s

These two waves (from examples 1 and 2), both travel at roughly the same speed, 336 m/s  and  337.5 m/s , this is because the speed of sound is fixed for any wave of any frequency travelling through air.

Example 3
The sound wave in example 2 passing through water, the frequency remains unchanged but the wavelength increases to  39.9 cm , what is the new speed of the wave?v = ?
f = 3750 Hz
\lambda = 39.9 cm = 0.399 m

v = f \lambda
v = 3750 \times 0.09
v = 1498 m/s

The speed of the sound wave increases from 337.5 m/s to 1498 m/s when travelling from air into water. This is because water is more dense and so the particles do not need to travel as far during each vibration before it strikes the next particle.

The more dense the medium, the faster sound will travel.

Examples using this equation

Use your understanding of the wave speed equation to answer the following questions (the answers are below the questions so you can try them and see if you are doing them correctly):

  1. Light travels at 3 \times 10^{8} m/s in air, the frequency of the a particular colour was measured to be  4.3 \times 10^{14} Hz what is the wavelength of the light?
  2. Sound travelling through an aluminium pipe with a wavelength of  44 cm , and frequency of  14.4 kHz has what speed?
  3. A laser beam is shone through some water, in the water it has a wavelength of  6.75 \times 10^{-7} m and a frequency of   3.33 \times 10^{14} Hz , what speed does light travel through water at? Does light slow down or speed up when it goes from air into water?

Answers

  1. 6.98 \times 10^{-7} m
  2. 6336 m/s
  3. 2.25 \times 10^{8} m/s
    The light has gone from 3 \times 10^{8} m/s in air to  2.25 \times 10^{8} m/s in water, it therefore slows down.

Further Reading