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In this Physics tutorial, you will learn:

- What are the factors affecting the energy of waves?
- What is the equation of kinetic energy of waves?
- The same for the potential energy of waves.
- How do these energies compare?
- How can we calculate the total energy of waves?
- What is power of a wave?
- How can we calculate the power of a mechanical wave?

Do you know any example of waves carrying energy?

Have you ever heard of seismic waves? What do they cause?

Why it is not good to stay for long periods exposed to direct sunlight? Can you explain your opinion in terms of electromagnetic waves?

How do water waves chew up beaches?

How does loud sound damage the hearing?

All the above examples involve energy (and therefore power) of various kinds of waves which we will discuss in this tutorial. Also, the mathematical procedure to obtain the equations of waves energy and power will be explained.

As stated in the previous tutorials, waves do not carry matter but only energy. This is because in all waves (although this is more visible in transverse waves) a particle oscillates around an equilibrium position giving a zero resultant displacement, while the wave spreads in a certain direction. Look at the figure:

Energy of waves depends on two factors: **amplitude** and **frequency**. This is because when a particle of a wave as the one shown in the above figure oscillates up and down, its gravitational potential energy depends on the amplitude, i.e. how far it displaces from the equilibrium position. Therefore, a greater amplitude means a greater gravitational potential energy for this particle when it reaches the maximum position.

On the other hand, it is a known fact that kinetic energy depends on the moving speed (KE = m × v^{2} / 2) and the latter depends on the wave frequency if wavelength is taken as constant (v = λ × f=>v ~ f for constant λ).

For example, the strength of seismic waves, which cause up and down oscillations of the Earth surface during earthquakes, depends on the amplitude of oscillations. Stronger the earthquake, greater the amplitude of seismic waves. On the other hand, the energy of EM waves depends on their frequency. Thus, higher the frequency of EM waves, higher their penetrating ability and therefore, greater the energy such waves carry with them.

Now, let's find a formula for the energy and then for the power of waves based on the actual knowledge on waves and their properties.

Consider an oscillating spring of mass m as shown in the figure.

We chose such a spring as its motion is both a combination of transverse and longitudinal waves, so the study of energy will be more complete.

The string oscillated up and down, so its kinetic energy is

KE = (m × v^{2}_{y})/2

In the previous tutorial, we have shown that the oscillating speed v_{y} of a wave is

v_{y} = *dy**/**dt*

=*d[A × sin(k × x - ω × t)]**/**dt*

= -A × ω × cos(k × x - ω × t)

=

= -A × ω × cos(k × x - ω × t)

Hence, we obtain for the kinetic energy of the oscillating spring:

KE = *m × [-A × ω × cos(k × x - ω × t) ]*^{2}*/**2*

=*m × A*^{2} × ω^{2} × cos^{2} (k × x - ω × t)*/**2*

=

As we know, cosine values vary from -1 to + 1 but when they are raised in power two, they become always positive. Hence, they vary from 0 to 1. This means the average value of all cosines at power two is 1/2. Therefore, we obtain for the kinetic energy of spring:

KE = *1**/**2* × *m × A*^{2} × ω^{2}*/**2*

=*m × A*^{2} × ω^{2}*/**4*

=

Also, we know that the potential energy PE of an oscillating spring is calculated by the equation

PE = *k × y*^{2}*/**2*

and giving that

y(x,t) = A × sin(k × x - ω × t)

we obtain for the potential energy of spring:

PE = *k × [A × sin(k × x - ω × t) ]*^{2}*/**2*

=*k × A*^{2} × sin^{2} (k × x - ω × t)*/**2*

=

In the tutorial "Simple Harmonic Motion", we have explained that the relationship between spring constant k, mass m and angular frequency ω is

ω^{2} = *k**/**m*

Thus,

k = ω^{2} × m

Substituting this value of k in the equation of potential energy, we obtain

PE = *ω*^{2} × m × A^{2} × sin^{2} (k × x - ω × t)*/**2*

Like in the cosine function, the square of sine function is equal to ** 1/2** as well. Remember the fundamental equation of trigonometry

cos^{2} x + sin^{2} x = 1

This means each of terms is equal to ** 1/2** because

PE = *1**/**2* × *ω*^{2} × m × A^{2}*/**2*

=*m × A*^{2} × ω^{2}*/**4*

=

This is the same result as the result obtained for kinetic energy. Hence, we can write for the total (mechanical) energy of the oscillating spring (here we have a spring but this approach can be applied in all situations involving waves):

Total Energy = ME

= KE + PE

=*m × A*^{2} × ω^{2}*/**4* + *m × A*^{2} × ω^{2}*/**4*

=*2 × m × A*^{2} × ω^{2}*/**4*

=*m × A*^{2} × ω^{2}*/**2*

= KE + PE

=

=

=

A 150 g rope shakes up and down as shown in the figure.

If the amplitude of wave caused by the shaking process is 20 cm and the up and down movement of the person who shakes the rope occurs every 2 seconds, what is the energy delivered by the rope's wave?

Clues:

m = 150 g = 0.150 kg

A = 20 cm = 0.20 m

f = 1 oscillation / 2 s = 0.5 oscillation/ second = 0.5 Hz

E = ?

First, let's calculate the angular frequency ω. Thus, giving that

ω = 2 × π × f

we obtain

ω = 2 × 3.14 × 0.5

= 3.14 rad/s

= 3.14 rad/s

Now, let's apply the formula of wave's energy. Thus,

E = *m × A*^{2} × ω^{2}*/**2*

=*0.150 × 0.20*^{2} × 3.14^{2}*/**2*

= 0.0296 J

=

= 0.0296 J

As expected, the energy delivered by such a wave is small. Otherwise, we would use shaking ropes as a source of energy.

As we know, power is the work done (or the energy delivered) by a system in the unit of time. In this regard, it would be very easy to find a formula for power of a wave giving that

E = *m × A*^{2} × ω^{2}*/**2*

Thus, we would simply divide this value by time t to obtain the formula of power, i.e.

P = *m × A*^{2} × ω^{2}*/**2t*

However, this is not always possible as in most cases we don't have any info about the time, or worse, the time is infinity as the wave is standing. Thus, we must find another formula for power of waves that is independent from the time.

Let's consider again the rope of the previous exercise. We take a small piece of it, with mass Δm and length Δx ash shown in the figure.

We define the linear mass density μ as the mass per unit length.

μ = *∆m**/**∆x*

Thus,

∆m = μ × ∆x

Given that the wave performs uniform motion, we have

∆x = v × ∆t

Hence,

∆m = μ × v × ∆t

If we consider the entire rope, we will obtain for the mass m

m = μ × v × t

Substituting this equation in the equation we previously written for power of wave, we obtain

P = *m × A*^{2} × ω^{2}*/**2t*

=*μ × v × t × A*^{2} × ω^{2}*/**2t*

=*μ × v × A*^{2} × ω^{2}*/**2*

=

=

This expression is independent from the time t as this quantity does not appear in the formula of wave's power anymore.

A rope of mass density 0.05 kg/m shakes at 30 cm amplitude. The wave produced, propagates at 2 m/s along the rope. The wave is produced by shaking the rope at 1.2 cycles per second. What is the power delivered by the wave?

Clues:

μ = 0.05 kg/m

A = 30 cm = 0.30 m

v = 2 m/s

f = 1.2 cycles/s = 1.2 Hz

P = ?

Let's calculate the angular frequency ω first. Thus,

ω = 2 × π × f

= 2 × 3.14 × 1.2

= 7.536 rad/s

= 2 × 3.14 × 1.2

= 7.536 rad/s

Thus, the power of this wave is

P = *μ × v × A*^{2} × ω^{2}*/**2*

=*0.05 × 2 × 0.30*^{2} × 7.536^{2}*/**2*

= 0.256 W

=

= 0.256 W

Waves do not carry matter but only energy. This is because in all waves (although this is more visible in transverse waves) a particle oscillates around an equilibrium position giving a zero resultant displacement, while the wave spreads in a certain direction.

Energy of waves depends on two factors: **amplitude** and **frequency**. This is because when a particle of a wave oscillates up and down, its gravitational potential energy depends on the amplitude, i.e. how far it displaces from the equilibrium position. Therefore, a greater amplitude means a greater gravitational potential energy for this particle when it reaches the maximum position.

On the other hand, it is a known fact that kinetic energy depends on the moving speed (KE = m × v^{2} / 2) and the latter depends on the wave frequency if wavelength is taken as constant (v = λ × f=>v ~ f for constant λ).

The kinetic energy of the oscillating spring is

KE = *m × [-A × ω × cos(k × x - ω × t) ]*^{2}*/**2*

=*m × A*^{2} × ω^{2} × cos^{2} (k × x - ω × t)*/**2*

=

Cosine values vary from -1 to + 1 but when they are raised in power two, they become always positive. Hence, they vary from 0 to 1. This means the average value of all cosines at power two is 1/2. Therefore, we obtain for the kinetic energy of spring:

KE = *1**/**2* × *m × A*^{2} × ω^{2}*/**2*

=*m × A*^{2} × ω^{2}*/**4*

=

Also, we know that the potential energy PE of an oscillating spring is calculated by the equation

PE = *k × y*^{2}*/**2*

and giving that

y(x,t) = A × sin(k × x - ω × t)

we obtain for the potential energy of spring:

PE = *k × [A × sin(k × x - ω × t) ]*^{2}*/**2*

=*k × A*^{2} × sin^{2} (k × x - ω × t)*/**2*

=

Given that

ω^{2} = *k**/**m*

Thus,

k = ω^{2} × m

Substituting this value of k in the equation of potential energy, we obtain

PE = *ω*^{2} × m × A^{2} × sin^{2} (k × x - ω × t)*/**2*

Like in the cosine function, the square of sine function is equal to ** 1/2** as well. Remember the fundamental equation of trigonometry

cos^{2} x + sin^{2} x = 1

This means each of terms is equal to ** 1/2** because

PE = *1**/**2* × *ω*^{2} × m × A^{2}*/**2*

=*m × A*^{2} × ω^{2}*/**4*

=

This is the same result as the result obtained for kinetic energy. Hence, we can write for the total (mechanical) energy of the oscillating spring (here we have a spring but this approach can be applied in all situations involving waves):

Total Energy = ME

= KE + PE

=*m × A*^{2} × ω^{2}*/**4* + *m × A*^{2} × ω^{2}*/**4*

=*m × A*^{2} × ω^{2}*/**2*

= KE + PE

=

=

Given that power is the work done (or the energy delivered) by a system in the unit of time, we obtain for power of waves:

P = *E**/**t*

=*m × A*^{2} × ω^{2}*/**2t*

=*μ × v × t × A*^{2} × ω^{2}*/**2t*

=*μ × v × A*^{2} × ω^{2}*/**2*

=

=

=

where μ is the linear mass density in kg/m.

This expression is independent from the time t as this quantity does not appear in the formula of wave's power anymore.

*1. A 200 g rope is oscillating as shown in the figure. *

What is the kinetic energy of transferred by the rope?

- 0.006 J
- 0.003 J
- 60 J
- 30 J

**Correct Answer: B**

*2. What is the total energy transferred by a 40 m3 water wave if it is 60 cm high? The wave is moving at 1 m/s and its wavelength is 2 m. Take the density of water equal to 1000 kg/m3. *

- 70 989 J
- 35 495 J
- 7 099 J
- 11 304 J

**Correct Answer: A**

*3. A 100 g rope shakes as shown in the figure. *

What is the power delivered by the rope if it is 2 m long when stretched?

- 105 W
- 104 W
- 10-4 W
- 10-5 W

**Correct Answer: D**

We hope you found this Physics tutorial "Energy and Power of Waves" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Waves with our Physics tutorial on Interference of Waves .

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