You are here: Home Speed of a transverse wave

## Speed of a transverse wave

Definition 1: Wave speed

Wave speed is the distance a wave travels per unit time.

Quantity: Wave speed ($$v$$)         Unit name: metre per second         Unit symbol: m·s−1

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

$v=distance travelled time taken=λ T$(1)

However, $f=1 T$. Therefore, we can also write:

$v=λ T=λ·1 T=λ·f$(2)

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

• $v=$ speed in m·s−1

• $λ=$ wavelength in m

• $f=$ frequency in Hz

Wave equation:

$v=f·λorv=λ T$(3)

## Example 1: Speed of a transverse wave I

### Question

When a particular string is vibrated at a frequency of 10 Hz, a transverse wave of wavelength 0,25 m is produced. Determine the speed of the wave as it travels along the string.

#### Determine what is given and what is required

• frequency of wave: $f=$ 10 Hz

• wavelength of wave: $λ=$ 0,25 m

We are required to calculate the speed of the wave as it travels along the string.

All quantities are in SI units.

#### Determine how to approach the problem

We know that the speed of a wave is:

$v=f·λ$(4)

and we are given all the necessary quantities.

#### Substituting in the values

(5)

The wave travels at 2,5 m·s−1 along the string.

## Example 2: Speed of a transverse wave II

### Question

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 cm. If the cork is 2 m from the edge of the pool, how long does it take a ripple passing the cork to reach the edge?

#### Determine what is given and what is required

We are given:

• frequency of wave: $f=$ 1 Hz

• wavelength of wave: $λ=$ 20 cm

• distance of cork from edge of pool: $D=$ 2 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.

#### Determine how to approach the problem

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

$t=D v(fromv=D t)$(6)

We know that

$v=f·λ$(7)

Therefore,

$t=D f·λ$(8)

#### Convert wavelength to SI units

$20cm=0,2m$(9)

#### Solve the problem

$t=D f·λ=2m (1Hz)(0,2m)=2m (1s -1 )(0,2m)=10s$(10)

A ripple passing the leaf will take 10 s to reach the edge of the pool.

Video on waves

## Exercise 1: Waves

When the particles of a medium move perpendicular to the direction of the wave motion, the wave is called a _____________ wave.

transverse

A transverse wave is moving downwards. In what direction do the particles in the medium move?

The particles will move sideways, perpendicular to the direction of the wave.

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            .

a) D

b) B

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, crest to crest distance (horizontal) = 5 cm

The transverse waves are given in the figure above.  Wave 1 is the solid black line. Wave 2 is the dashed red line that has an amplitude of 1.5 cm and a wavelength of 5 cm.

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.

a) Twice the amplitude is 2 cm.

b) Half the amplitude is 0,5 cm.

c) $v=f\lambda =\frac{2}{2}f\lambda =\left(2f\right)\left(\frac{\lambda }{2}\right)$ thus if you have twice the frequency we have half the wavelength which is 1 cm.

d)$v=f\lambda =\frac{2}{2}f\lambda =\left(\frac{f}{2}\right)\left(2\lambda \right)$ thus if you have half the frequency we have twice the wavelength which is 4 cm.

e)  Twice the wavelength is 4 cm.

f) Half the wavelength is 1 cm.

g) $v=\frac{\lambda }{T}=\frac{2\lambda }{2T}$ thus if you have twice the period we have twice the wavelength which is 4 cm.

h) $v=\frac{\lambda }{T}=\frac{2\lambda }{2T}=\left(\frac{1}{2}\lambda \right)÷\left(\frac{1}{2}T\right)$ thus if you have twice the period we have twice the wavelength which is 4 cm.

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.

a)

 $T=\frac{1}{f}$ $T=\frac{1}{15}$ $T=0,067\text{s}$

The period of the wave is 0,067 s

b) The horizontal distance from peak to trough is half the wavelength, so the wavelength is 5 cm.

The speed is:

 $v=\lambda \cdot f$ $v=\left(5\text{cm}\right)\left(15{\text{s}}^{-1}\right)$ $v=75\text{cm}\cdot {\text{s}}^{-1}$ $v=0,75×{10}^{-2}\text{m}\cdot {\text{s}}^{-1}$

A fly flaps its wings back and forth 200 times each second. Calculate the period of a wing flap.

The frequency of the wing flap is 200 Hz. (In one second the fly flaps its wings 200 times).

The period is:

 $T=\frac{1}{f}$ $T=\frac{1}{200}$ $T=0,005s$

As the period of a wave increases, the frequency increases/decreases/does not change.

The frequency will decrease. The frequency is inversely related to the period and so when the period increases, the frequency will decrease.

Calculate the frequency of rotation of the second hand on a clock.

In one minute, the second hand moves 60 times, in other words it moves at a speed of one unit per second in one revolution for 60 units. And the revolution has a period of 60 seconds.

$f=\frac{1}{T}=\frac{1}{60}=0,017\text{Hz}$

Microwave ovens produce radiation with a frequency of 2450 MHz (1 MHz = 106 Hz) and a wavelength of 0,122 m. What is the wave speed of the radiation?

 $v=\lambda \cdot f$ $v=\left(0,122\right)\left(2450×{10}^{6}\right)$

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.

a) There are many sets of points, they are: (C,K),  (G, O) , (E, M), (A, I),  (I, Q), (B, J), (D, L), (F, N) and (H, P).

Two points will be in phase if they occur in the same position on a wave, so the two peaks a and the two troughs are in phase.

b) Points out of phase would be any pair of points not mentioned in (a). For example, A and F, B and Q, etc.

c) C and K or G and O. As well as any other two points that are in phase as in (a).

A wavelength is the distance from one peak to the next or from one trough to the next.

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.

The wavelength is 4 m, since this is the distance between successive crests. The period is 2 sec since four waves pass in 8 seconds.

Therefore the speed is:

 $v=\frac{\lambda }{T}$ $v=\frac{4}{2}$