Physics A Level
11.4 Potentiometer circuits 12.4 Wave speed
Physics A Level
11.4 Potentiometer circuits 12.4 Wave speed
The speed with which energy is transmitted by a wave is known as the wave speed v. This is measured in $m\,{s^{ - 1}}$. The wave speed for sound in air at atmospheric pressure of ${10^5}$ Pa and a temperature of ${0^ \circ }C$ is about $330\,m\,{s^{ - 1}}$, while for light in a vacuum it is almost $300000000\,m\,{s^{ - 1}}$.
The wave equation
An important equation connecting the speed v of a wave with its frequency, f and wavelength, $\lambda $ can be determined as follows. We can find the speed of the wave using:
$speed = \frac{{distance}}{{time}}$
A wave will travel a distance of one whole wavelength, $\lambda $ in a time equal to one period, T. So:
$Wave\,speed = \frac{{wavelength}}{{period}}$ or $v = \frac{\lambda }{T}$
$v = \frac{1}{T} \times \lambda $
However, $f = \frac{1}{T}$ and so:
$v = f \times \lambda $
where v is the speed of the wave, f is the frequency and $\lambda $ is the wavelength.
A numerical example may help to make this clear. Imagine a wave of frequency $5 Hz$ and wavelength $3 m$ going past you. In $1 s$, five complete wave cycles, each of length 3 m, go past. So the total length of the waves going past in $1 s$ is $15 m$. The distance travelled by the wave per second is its speed, therefore the speed of the wave is $v = f15\,m\,{s^{ - 1}}$.
You can see that, for a given speed of wave, the greater the wavelength, the smaller the frequency (and the smaller the wavelength, the greater the frequency). This means, that for a constant wave speed, the wavelength is inversely proportional to the frequency. The speed of sound in air is constant (for a given temperature and pressure). The wavelength of sound can be made smaller by increasing the frequency of the source of sound.
Table 12.1 gives typical values of speed v, frequency f and wavelength $\lambda $ for some mechanical waves. You can check for yourself that $v = f\lambda $ is valid.
| Water waves in a ripple tank | Sound waves in air | Waves on a toy spring | |
| Speed $v/m\,{s^{ - 1}}$ | about 0.12 | 330 | about 1 |
| Frequency f / Hz | about 6 | 20 to 20 000 (limits of human hearing) | about 2 |
| Wavelength $\lambda /m$ | about 0.2 | 16.5 to 0.0165 | about 0.5 |
WORKED EXAMPLE
2) Middle C on a piano tuned to concert pitch should have a frequency of $264 Hz$ (Figure 12.10). If the speed of sound is $330\,m\,{s^{ - 1}}$, calculate the wavelength of the sound produced from this note.
Step 1: We rearrange the wave equation to the form:
$\lambda = \frac{v}{f}$
Step2: Substituting the values we get:
$\lambda = \frac{{330}}{{264}} = 1.25m$
The wavelength, $\lambda $ is $1.25 m$.
Questions
6) Sound is a mechanical wave that can be transmitted through a solid.
Calculate the frequency of sound of wavelength $0.25 m$ that travels through steel at a speed of $5060\,m\,{s^{ - 1}}$
7) A cello string vibrates with a frequency of $64 Hz$.
Calculate the speed of the transverse waves on the cello string given that the wavelength is $140 cm$.
8) An oscillator is used to send a transverse wave along a stretched string. The wavelength of the wave is $5.0 cm$ when the frequency of the oscillator is $30 Hz$.
For this wave, calculate:
a: its frequency
b: its speed.
9) Copy and complete Table 12.2. (You may assume that the speed of radio waves is $3.00 \times {10^8}\,m\,{s^{ - 1}}$.)
| Station | Wavelength / m | Frequency / MHz |
| Radio A (FM) | 97.6 | |
| Radio B (FM) | 94.6 | |
| Radio B (LW) | 1515 | |
| Radio C (MW) | 693 |