# Introduction and key concepts

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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 this time 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 (s).

The period (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.

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 $\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}s$ , then the period of the wave is $\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}s$ . Therefore, the frequency of the wave is:

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

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

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 $0,1\phantom{\rule{2pt}{0ex}}s$ .

## Speed of a transverse wave

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

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

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:

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

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 $m·s{}^{-1}$
• $\lambda =$ wavelength in $m$
• $f=$ 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 $0,25\phantom{\rule{2pt}{0ex}}m$ is produced. Determine the speed of the wave as it travels along the string.

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

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 $2,5\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ along the string.

A cork on the surface of a swimming pool bobs up and down once every second on some ripples. The ripples have a wavelength of $20\phantom{\rule{2pt}{0ex}}\mathrm{cm}$ . If the cork is $2\phantom{\rule{2pt}{0ex}}m$ from the edge of the pool, how long does it take a ripple passing the cork to reach the edge?

1. We are given:

• frequency of wave: $f=1\phantom{\rule{2pt}{0ex}}\mathrm{Hz}$
• wavelength of wave: $\lambda =20\phantom{\rule{2pt}{0ex}}\mathrm{cm}$
• distance of cork from edge of pool: $d\phantom{\rule{0.166667em}{0ex}}=2\phantom{\rule{2pt}{0ex}}m$

We are required to determine the time it takes for a ripple to travel between the cork and the edge of the pool.

The wavelength is not in SI units and should be converted.

2. The time taken for the ripple to reach the edge of the pool is obtained from:

$t=\frac{d}{v}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(\text{from}\phantom{\rule{4pt}{0ex}}v=\frac{d}{t}\right)$

We know that

$v=f·\lambda$

Therefore,

$t=\frac{d}{f·\lambda }$
3. $20\phantom{\rule{0.166667em}{0ex}}\mathrm{cm}=0,2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}$
4. $\begin{array}{ccc}\hfill t& =& \frac{d}{f·\lambda }\hfill \\ & =& \frac{2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}}{\left(1\phantom{\rule{0.277778em}{0ex}}\mathrm{Hz}\right)\left(0,2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\right)}\hfill \\ & =& 10\phantom{\rule{0.166667em}{0ex}}\mathrm{s}\hfill \end{array}$
5. A ripple passing the leaf will take $10\phantom{\rule{2pt}{0ex}}s$ to reach the edge of the pool.

The following video provides a summary of the concepts covered so far.

## Waves

1. When the particles of a medium move perpendicular to the direction of the wave motion, the wave is called a $.........$ wave.
2. A transverse wave is moving downwards. In what direction do the particles in the medium move?
3. Consider the diagram below and answer the questions that follow:
1. the wavelength of the wave is shown by letter .
2. the amplitude of the wave is shown by letter .
4. Draw 2 wavelengths of the following transverse waves on the same graph paper. Label all important values.
1. Wave 1: Amplitude = 1 cm, wavelength = 3 cm
2. Wave 2: Peak to trough distance (vertical) = 3 cm, peak to peak distance (horizontal) = 5 cm
5. You are given the transverse wave below. Draw the following:
1. A wave with twice the amplitude of the given wave.
2. A wave with half the amplitude of the given wave.
3. A wave travelling at the same speed with twice the frequency of the given wave.
4. A wave travelling at the same speed with half the frequency of the given wave.
5. A wave with twice the wavelength of the given wave.
6. A wave with half the wavelength of the given wave.
7. A wave travelling at the same speed with twice the period of the given wave.
8. A wave travelling at the same speed with half the period of the given wave.
6. A transverse wave travelling at the same speed with an amplitude of 5 cm has a frequency of 15 Hz. The horizontal distance from a crest to the nearest trough is measured to be 2,5 cm. Find the
1. period of the wave.
2. speed of the wave.
7. A fly flaps its wings back and forth 200 times each second. Calculate the period of a wing flap.
8. As the period of a wave increases, the frequency increases/decreases/does not change.
9. Calculate the frequency of rotation of the second hand on a clock.
10. Microwave ovens produce radiation with a frequency of 2 450 MHz (1 MHz = ${10}^{6}$  Hz) and a wavelength of 0,122 m. What is the wave speed of the radiation?
11. Study the following diagram and answer the questions:
1. Identify two sets of points that are in phase.
2. Identify two sets of points that are out of phase.
3. Identify any two points that would indicate a wavelength.
12. Tom is fishing from a pier and notices that four wave crests pass by in 8 s and estimates the distance between two successive crests is 4 m. The timing starts with the first crest and ends with the fourth. Calculate the speed of the wave.

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can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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