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Wize University Physics Textbook (Master) > Waves: Mechanical

Power and Intensity of Waves

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Transmitted Power and Intensity


There is energy transmitted by a wave. For a mechanical wave on a string, we can calculate the rate at which this energy is transmitted in space. This rate of energy transmission is known as power and it has the following form:

P=ΔEΔt=μvω2A2cos⁡2(kx−ωt)P=\frac{\Delta E}{\Delta t}=\mu v\omega^2A^2\cos^2\left(kx-\omega t\right)


  • PPis the power in watt
  • EEis the energy in jj
  • μ\mu is the mass density in kg/mkg/m
  • vv is the wave speed in m/sm/s
  • AAis the amplitude in mm
  • kkis the wave number in rad/mrad/m
  • ω\omegais the angular frequency in rad/srad/s
  • FFis the tension in NN


Average Power Transmitted

Most of the time, we are only interested to know the average power transmitted by a wave:


Pavg=12μvω2A2=12μFω2A2\boxed{P_{avg}=\frac{1}{2}\mu v\omega^2A^2=\frac{1}{2}\sqrt{\mu F}\omega^2A^2}







The maximum power in a sinusoidal wave is:

Pmax⁡=μFω2A2=2Pavg\boxed{P_{\max}=\sqrt{\mu F}\omega^2A^2=2P_{avg}}



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Intensity

Intensity of a wave is defined for 3-dimensional waves and is the average power per unit area.


I=PA\boxed{I=\dfrac{P}{A}}


Intensity of a wave at a distance rr from the source could be written as:

I=P4πr2\boxed{I=\dfrac{P}{4\pi r^2}}
























Watch Out!
Note that intensity drops quickly by 1/r21/r^2when you get farther away from the source.

Wize Tip
For two different distances from the same source, we have the inverse-square law for intensity


I1I2=r22r12\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}

this equation is very useful in solving intensity problems. But be careful using it because it only works for the same sources.





Example: Intensity of Sunlight

When they reach the earth, light waves emitted by the sun average to a total power of 1.3621.362 kW/m2.

a) Given that the orbital radius of the Earth is 1.496×1081.496\times10^8km, find the total power output of the Sun.
b) Find the intensity of sunlight that reaches an asteroid 5.45.4 times farther away from the Sun than the Earth.

Part a)

The total power output of the sun is spread out over an imaginary sphere with the radius equal to the orbital radius of the earth.
We're given the intensity (power / unit area) and distance, so let's use:

I=P4πr2   →   P=4πr2II=\dfrac{P}{4\pi r^2} \ \ \ \to \ \ \ P=4\pi r^2I
=4π(1.496×1011)2(1362)=4\pi (1.496\times10^{11})^2(1362)
=3.83×1026=3.83\times10^{26} (W/m2)

Part b)

We know that rasteroid=5.4×rearthr_{asteroid}=5.4 \times r_{earth}

To compare the intensities, we use:

IearthIasteroid=rasteroid2rearth2\dfrac{I_{earth}}{I_{asteroid}}=\dfrac{r_{asteroid}^2}{r_{earth}^2}

=(5.4×rearth)2rearth2=\dfrac{(5.4 \times r_{earth})^2}{r_{earth}^2}

=5.42=5.4^2

=29.16=29.16


This means that Iearth=29.16×Iasteroid{I_{earth}}=29.16 \times {I_{asteroid}}, or:


Iasteroid=Iearth29.16{I_{asteroid}}=\dfrac{{I_{earth}}}{29.16}

=136229.16=\dfrac{1362}{29.16}

=46.7= 46.7 (W/m2)

Practice: Power of a Traveling Wave


A string is known to have a linear density of 1010 g/cm. A wave traveling down on it has an amplitude of 55 cm, oscillation period of 0.50.5 s, and a wavelength of 1010 cm. What is the average power of this traveling wave?