# Energy and Power of Waves

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11.2Energy and Power of Waves

In these revision notes for Energy and Power of Waves, we cover the following key points:

• 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?

## Energy and Power of Waves Revision Notes

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 × v2 / 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 × A2 × ω2 × cos2 (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 × A2 × ω2/2
= m × A2 × ω2/4

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

PE = k × y2/2

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 × A2 × sin2 (k × x - ω × t)/2

Given that

ω2 = k/m

Thus,

k = ω2 × m

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

PE = ω2 × m × A2 × sin2 (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

cos2 x + sin2 x = 1

This means each of terms is equal to 1/2 because 1/2 + 1/2 = 1. Therefore, we have

PE = 1/2 × ω2 × m × A2/2
= m × A2 × ω2/4

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 × A2 × ω2/4 + m × A2 × ω2/4
= m × A2 × ω2/2

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 × A2 × ω2/2t
= μ × v × t × A2 × ω2/2t
= μ × v × A2 × ω2/2

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.

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