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  5. Energy and power of a wave

16.4 Energy and power of a wave  (Page 2/6)

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Figure shows a box to the left, labeled string vibrator. A string is attached to this and forms a transverse wave that propagates towards the right with velocity v subscript w. A small portion of the string is highlighted and is labeled delta m.
A string vibrator is a device that vibrates a rod. A string is attached to the rod, and the rod does work on the string, driving the string up and down. This produces a sinusoidal wave in the string, which moves with a wave velocity v . The wave speed depends on the tension in the string and the linear mass density of the string. A section of the string with mass Δ m oscillates at the same frequency as the wave.

The total mechanical energy of the wave is the sum of its kinetic energy and potential energy. The kinetic energy K = 1 2 m v 2 of each mass element of the string of length Δ x is Δ K = 1 2 ( Δ m ) v y 2 , as the mass element oscillates perpendicular to the direction of the motion of the wave. Using the constant linear mass density, the kinetic energy of each mass element of the string with length Δ x is

Δ K = 1 2 ( μ Δ x ) v y 2 .

A differential equation can be formed by letting the length of the mass element of the string approach zero,

d K = lim Δ x → 0 1 2 ( μ Δ x ) v y 2 = 1 2 ( μ d x ) v y 2 .

Since the wave is a sinusoidal wave with an angular frequency ω , the position of each mass element may be modeled as y ( x , t ) = A sin ( k x − ω t ) . Each mass element of the string oscillates with a velocity v y = ∂ y ( x , t ) ∂ t = − A ω cos ( k x − ω t ) . The kinetic energy of each mass element of the string becomes

d K = 1 2 ( μ d x ) ( − A ω cos ( k x − ω t ) ) 2 , = 1 2 ( μ d x ) A 2 ω 2 cos 2 ( k x − ω t ) .

The wave can be very long, consisting of many wavelengths. To standardize the energy, consider the kinetic energy associated with a wavelength of the wave. This kinetic energy can be integrated over the wavelength to find the energy associated with each wavelength of the wave:

d K = 1 2 ( μ d x ) A 2 ω 2 cos 2 ( k x ) , ∫ 0 K λ d K = ∫ 0 λ 1 2 μ A 2 ω 2 cos 2 ( k x ) d x = 1 2 μ A 2 ω 2 ∫ 0 λ cos 2 ( k x ) d x , K λ = 1 2 μ A 2 ω 2 [ 1 2 x + 1 4 k sin ( 2 k x ) ] 0 λ = 1 2 μ A 2 ω 2 [ 1 2 λ + 1 4 k sin ( 2 k λ ) − 1 4 k sin ( 0 ) ] , K λ = 1 4 μ A 2 ω 2 λ .

There is also potential energy associated with the wave. Much like the mass oscillating on a spring, there is a conservative restoring force that, when the mass element is displaced from the equilibrium position, drives the mass element back to the equilibrium position. The potential energy of the mass element can be found by considering the linear restoring force of the string, In Oscillations , we saw that the potential energy stored in a spring with a linear restoring force is equal to U = 1 2 k s x 2 , where the equilibrium position is defined as x = 0.00 m . When a mass attached to the spring oscillates in simple harmonic motion, the angular frequency is equal to ω = k s m . As each mass element oscillates in simple harmonic motion, the spring constant is equal to k s = Δ m ω 2 . The potential energy of the mass element is equal to

Δ U = 1 2 k s x 2 = 1 2 Δ m ω 2 x 2 .

Note that k s is the spring constant and not the wave number k = 2 π λ . This equation can be used to find the energy over a wavelength. Integrating over the wavelength, we can compute the potential energy over a wavelength:

d U = 1 2 k s x 2 = 1 2 μ ω 2 x 2 d x , U λ = 1 2 μ ω 2 A 2 ∫ 0 λ cos 2 ( k x ) d x = 1 4 μ A 2 ω 2 λ .

The potential energy associated with a wavelength of the wave is equal to the kinetic energy associated with a wavelength.

The total energy associated with a wavelength is the sum of the potential energy and the kinetic energy:
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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