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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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