# 0.4 Transverse waves  (Page 3/10)

 Page 3 / 10

We then have an alternate definition of the wavelength as the distance between any two adjacent points which are in phase .

Wavelength of wave

The wavelength of a wave is the distance between any two adjacent points that are in phase.

Points that are not in phase, those that are not separated by a complete number of wavelengths, are called out of phase . Examples of points like these would be $A$ and $C$ , or $D$ and $E$ , or $B$ and $H$ in the Activity.

## Period and frequency

Imagine you are sitting next to a pond and you watch the waves going past you. First one peak arrives, then a trough, and then another peak. Suppose you measure the time taken between one peak arriving and then the next. This time will be the same for any two successive peaks passing you. We call thistime the period , and it is a characteristic of the wave.

The symbol $T$ is used to represent the period. The period is measured in seconds ( $\mathrm{s}$ ).

Period ( $\mathrm{T}$ )
The period ( $\mathrm{T}$ ) is the time taken for two successive peaks (or troughs) to pass a fixed point.

Imagine the pond again. Just as a peak passes you, you start your stopwatch and count each peak going past. After 1 second you stop the clock and stop counting. The number of peaks that you have counted in the 1 second is the frequency of the wave.

Frequency
The frequency is the number of successive peaks (or troughs) passing a given point in 1 second.

The frequency and the period are related to each other. As the period is the time taken for 1 peak to pass, then the number of peaks passing the point in 1 second is $\frac{1}{T}$ . But this is the frequency. So

$f=\frac{1}{T}$

or alternatively,

$T=\frac{1}{f}$

For example, if the time between two consecutive peaks passing a fixed point is $\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}\mathrm{s}$ , then the period of the wave is $\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}\mathrm{s}$ . Therefore, the frequency of the wave is:

$\begin{array}{ccc}\hfill f& =& \frac{1}{T}\hfill \\ & =& \frac{1}{\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}\mathrm{s}}\hfill \\ & =& 2\phantom{\rule{0.166667em}{0ex}}{\mathrm{s}}^{-1}\hfill \end{array}$

The unit of frequency is the Hertz ( $\mathrm{Hz}$ ) or ${\mathrm{s}}^{-1}$ .

What is the period of a wave of frequency $10\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$ ?

1. We are required to calculate the period of a $10\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$ wave.

2. We know that:

$T=\frac{1}{f}$
3. $\begin{array}{ccc}\hfill T& =& \frac{1}{f}\hfill \\ & =& \frac{1}{10\phantom{\rule{0.166667em}{0ex}}\mathrm{Hz}}\hfill \\ & =& 0,1\phantom{\rule{0.166667em}{0ex}}\mathrm{s}\hfill \end{array}$
4. The period of a $10\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$ wave is $0,1\phantom{\rule{2pt}{0ex}}\mathrm{s}$ .

## Speed of a transverse wave

In Motion in One Dimension , we saw that speed was defined as

$\mathrm{speed}=\frac{\mathrm{distance}\phantom{\rule{2pt}{0ex}}\mathrm{traveled}}{\mathrm{time}\phantom{\rule{2pt}{0ex}}\mathrm{taken}}$

The distance between two successive peaks is 1 wavelength, $\lambda$ . Thus in a time of 1 period, the wave will travel 1 wavelength in distance. Thus the speed of the wave, $v$ , is:

$v=\frac{\text{distance}\phantom{\rule{4.pt}{0ex}}\text{traveled}}{\text{time}\phantom{\rule{4.pt}{0ex}}\text{taken}}=\frac{\lambda }{T}$

However, $f=\frac{1}{T}$ . Therefore, we can also write:

$\begin{array}{ccc}\hfill v& =& \frac{\lambda }{T}\hfill \\ & =& \lambda ·\frac{1}{T}\hfill \\ & =& \lambda ·f\hfill \end{array}$

We call this equation the wave equation . To summarise, we have that $v=\lambda ·f$ where

• $v=$ speed in $\mathrm{m}·\mathrm{s}{}^{-1}$
• $\lambda =$ wavelength in $\mathrm{m}$
• $f=$ frequency in $\mathrm{Hz}$

When a particular string is vibrated at a frequency of $10\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$ , a transverse wave of wavelength $0,25\phantom{\rule{2pt}{0ex}}\mathrm{m}$ is produced. Determine the speed of the wave as it travels along the string.

• frequency of wave: $f=10\mathrm{Hz}$
• wavelength of wave: $\lambda =0,25\mathrm{m}$

We are required to calculate the speed of the wave as it travels along the string. All quantities are in SI units.

1. We know that the speed of a wave is:

$v=f·\lambda$

and we are given all the necessary quantities.

2. $\begin{array}{ccc}\hfill v& =& f·\lambda \hfill \\ & =& \left(10\phantom{\rule{0.277778em}{0ex}}\mathrm{Hz}\right)\left(0,25\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\right)\hfill \\ & =& 2,5\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \end{array}$
3. The wave travels at $2,5\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ along the string.

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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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or in general
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in general
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what is biological synthesis of nanoparticles
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