Contents

Energy in Waves: Intensity

The destruction caused by an earthquake in Port-au-Prince, Haiti. Some buildings are shown on two sides of a street. Two buildings are completely destroyed. Rescue people are seen around.

Figure 16.44 The destructive effect of an earthquake is palpable evidence of the energy carried in these waves. The Richter scale rating of earthquakes is related to both their amplitude and the energy they carry. (credit: Petty Officer 2nd Class Candice Villarreal, U.S. Navy)

All waves carry energy. The energy of some waves can be directly observed. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls.

Loud sounds pulverize nerve cells in the inner ear, causing permanent hearing loss. Ultrasound is used for deep-heat treatment of muscle strains. A laser beam can burn away a malignancy. Water waves chew up beaches.

The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements. Loud sounds have higher pressure amplitudes and come from larger-amplitude source vibrations than soft sounds. Large ocean breakers churn up the shore more than small ones. More quantitatively, a wave is a displacement that is resisted by a restoring force. The larger the displacement $x x size 12{x} {}$ , the larger the force $F = kx F = kx size 12{F= ital "kx"} {}$ needed to create it. Because work $W W size 12{W} {}$ is related to force multiplied by distance ( $Fx Fx size 12{ ital "Fx"} {}$ ) and energy is put into the wave by the work done to create it, the energy in a wave is related to amplitude. In fact, a wave’s energy is directly proportional to its amplitude squared because

W \propto Fx = {kx}^{2} . W \propto Fx = {kx}^{2} . size 12{W prop  ital "Fx"= ital "kx" rSup { size 8{2} } } {}

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The energy effects of a wave depend on time as well as amplitude. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight, for example, can be focused to burn wood. Earthquakes spread out, so they do less damage the farther they get from the source. In both cases, changing the area the waves cover has important effects. All these pertinent factors are included in the definition of intensity $I I size 12{I} {}$ as power per unit area:

I = \frac{P}{A} I = \frac{P}{A} size 12{I= {  {P}  over  {A} } } {}

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where $P P size 12{P} {}$ is the power carried by the wave through area $A A size 12{A} {}$ . The definition of intensity is valid for any energy in transit, including that carried by waves. The SI unit for intensity is watts per square meter ( ${W/m}^{2} size 12{"W/m" rSup { size 8{2} } } {}$ ). For example, infrared and visible energy from the Sun impinge on Earth at an intensity of $1300 1300 size 12{"1300"`"W/m" rSup { size 8{2} } } {}$ just above the atmosphere. There are other intensity-related units in use, too. The most common is the decibel. For example, a 90 decibel sound level corresponds to an intensity of $10^{- 3} 10^{- 3} size 12{"10" rSup { size 8{ - 3} } `"W/m" rSup { size 8{2} } } {}$ . (This quantity is not much power per unit area considering that 90 decibels is a relatively high sound level. Decibels will be discussed in some detail in a later chapter.

Example 16.9

Calculating intensity and power: How much energy is in a ray of sunlight?

The average intensity of sunlight on Earth’s surface is about $700 700 size 12{7"00"`"W/m" rSup { size 8{2} } } {}$ .

(a) Calculate the amount of energy that falls on a solar collector having an area of $0.500 0.500 size 12{0 "." "500"`"m" rSup { size 8{2} } } {}$ in $4.00 4.00 size 12{4 "." "00"`"h"} {}$ .

(b) What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own?

Strategy a

Because power is energy per unit time or $P = \frac{E}{t} P = \frac{E}{t} size 12{P= { {E} over {t} } } {}$ , the definition of intensity can be written as $I = \frac{P}{A} = \frac{E / t}{A} I = \frac{P}{A} = \frac{E / t}{A} size 12{I= { {P} over {A} } = { { {E} slash {t} } over {A} } } {}$ , and this equation can be solved for E with the given information.

Solution a

Begin with the equation that states the definition of intensity:
$I = \frac{P}{A} . I = \frac{P}{A} . size 12{I= { {P} over {A} } } {}$
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Replace $P P size 12{P} {}$ with its equivalent $E / t E / t size 12{E/t} {}$ :
$I = \frac{E / t}{A} . I = \frac{E / t}{A} . size 12{I= { { {E} slash {t} } over {A} } } {}$
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Solve for $E E size 12{P} {}$ :
$E = IAt . E = IAt . size 12{E= ital "IAt"} {}$
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Substitute known values into the equation:
$E = (700 {W/m}^{2}) (0.500 m^{2}) [(4.00 h) (3600 s/h)] . E = (700) (0.500) [(4.00) (3600)] . size 12{E= left ("700"" W/m" rSup { size 8{2} } right ) left (0 "." "500"" m" rSup { size 8{2} } right ) left [ left (4 "." "00"" h" right ) left ("3600"" s/h" right ) right ]} {}$
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Calculate to find $E E size 12{E} {}$ and convert units:
$5 . 04 \times 10^{6} J, 5 . 04 \times 10^{6}, size 12{5 "." "04" times "10" rSup { size 8{6} } "J"} {}$
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Discussion a

The energy falling on the solar collector in 4 h in part is enough to be useful—for example, for heating a significant amount of water.

Strategy b

Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.

Solution b

Take the ratio of intensities, which yields:
$\frac{I'}{I} = \frac{P' / A'}{P / A} = \frac{A}{A'} (The powers cancel because P' = P) . \frac{I'}{I} = \frac{P' / A'}{P / A} = \frac{A}{A'}$
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Identify the knowns:
$A = 200 A', A = 200 A', size 12{A="200"A'} {}$
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$\frac{I'}{I} = 200 . \frac{I'}{I} = 200 . size 12{ { {I rSup { size 8{'} } } over {I} } ="200"} {}$
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Substitute known quantities:
$I' = 200 I = 200 (700) . I' = 200 I = 200 (700) . size 12{I'="200"I="200" left ("700"`"W/m" rSup { size 8{2} } right )} {}$
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Calculate to find $I' I' size 12{I'} {}$ :
$I' = 1.40 \times 10^{5} . I' = 1.40 \times 10^{5} . size 12{ { {I}} sup { ' }=1 "." "40" times "10" rSup { size 8{5} } `"W/m" rSup { size 8{2} } } {}$
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Discussion b

Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.

Example 16.10

Determine the combined intensity of two waves: Perfect constructive interference

If two identical waves are spatially separated, each having an intensity of 1.00 W/m², interfere perfectly constructively in a particular location, what is the intensity of the wave at this particular location?

Strategy

We know from Superposition and Interference that when two identical waves, which have equal amplitudes $X X size 12{X} {}$ , interfere perfectly constructively, the resulting wave has an amplitude of $2 X 2 X size 12{2X} {}$ . Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.

Solution

Recall that intensity is proportional to amplitude squared.
Calculate the new amplitude:
$I' \propto {(X')}^{2} = {(2 X)}^{2} = {4 X}^{2} . I' \propto {(X')}^{2} = {(2 X)}^{2} = {4 X}^{2} . size 12{I rSup { size 8{'} } prop left (X rSup { size 8{'} } right ) rSup { size 8{2} } = left (2X right ) rSup { size 8{2} } =4X rSup { size 8{2} } } {}$
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Recall that the intensity of the old amplitude was:
$I \propto X^{2} . I \propto X^{2} . size 12{I rSup { size 8{'} } prop X rSup { size 8{2} } } {}$
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Take the ratio of new intensity to the old intensity. This gives:
$\frac{I'}{I} = 4 . \frac{I'}{I} = 4 . size 12{ { {I} over {I rSup { size 8{'} } } } =4} {}$
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Calculate to find $I' I' size 12{I'} {}$ :
$I' = 4 I = 4 . 00 . I' = 4 I = 4 . 00 . size 12{I'=4I=4 "." "00"`"W/m" rSup { size 8{2} } } {}$
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Discussion

The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of $1.00 1.00 size 12{1 "." "00"`"W/m" rSup { size 8{2} } } {}$ , yet their sum has an intensity of $4.00 4.00 size 12{4 "." "00"`"W/m" rSup { size 8{2} } } {}$ , which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is $4.00 4.00 size 12{4 "." "00"`"W/m" rSup { size 8{2} } } {}$ is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in Superposition and Interference suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out $1.00 1.00 size 12{1 "." "00"`"W/m" rSup { size 8{2} } } {}$ each, there will be places in the room where the intensity is $4.00 4.00 size 12{4 "." "00"`"W/m" rSup { size 8{2} } } {}$ , other places where the intensity is zero, and others in between. Figure 16.45 shows what this interference might look like. We will pursue interference patterns elsewhere in this text.

Two speakers are shown at the top of the figure at left and right side. Rarefactions are shown as dotted curves and compression as dark curves. The interference of the sound waves from these two speakers is shown. There are some red spots, showing constructive interference, are shown on the interfering waves.

Figure 16.45 These stereo speakers produce both constructive interference and destructive interference in the room, a property common to the superposition of all types of waves. The shading is proportional to intensity.

Check Your Understanding

Which measurement of a wave is most important when determining the wave's intensity?

Show solution

Solution

Amplitude, because a wave’s energy is directly proportional to its amplitude squared.