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Energy in Waves Intensity

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Energy in Waves Intensity
CHAPTER 16 | OSCILLATORY MOTION AND WAVES
While beats may sometimes be annoying in audible sounds, we will find that beats have many applications. Observing beats is a very useful way to
compare similar frequencies. There are applications of beats as apparently disparate as in ultrasonic imaging and radar speed traps.
Check Your Understanding
Imagine you are holding one end of a jump rope, and your friend holds the other. If your friend holds her end still, you can move your end up and
down, creating a transverse wave. If your friend then begins to move her end up and down, generating a wave in the opposite direction, what
resultant wave forms would you expect to see in the jump rope?
Solution
The rope would alternate between having waves with amplitudes two times the original amplitude and reaching equilibrium with no amplitude at
all. The wavelengths will result in both constructive and destructive interference
Check Your Understanding
Define nodes and antinodes.
Solution
Nodes are areas of wave interference where there is no motion. Antinodes are areas of wave interference where the motion is at its maximum
point.
Check Your Understanding
You hook up a stereo system. When you test the system, you notice that in one corner of the room, the sounds seem dull. In another area, the
sounds seem excessively loud. Describe how the sound moving about the room could result in these effects.
Solution
With multiple speakers putting out sounds into the room, and these sounds bouncing off walls, there is bound to be some wave interference. In
the dull areas, the interference is probably mostly destructive. In the louder areas, the interference is probably mostly constructive.
PhET Explorations: Wave Interference
Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern.
Figure 16.43 Wave Interference (http://cnx.org/content/m42249/1.5/wave-interference_en.jar)
16.11 Energy in Waves: Intensity
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 , the larger the force
579
580
CHAPTER 16 | OSCILLATORY MOTION AND WAVES
F = kx needed to create it. Because work W is related to force multiplied by distance ( 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 ∝ Fx = kx 2.
(16.74)
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 as power per unit area:
I=P
A
where
(16.75)
P is the power carried by the wave through area 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 ). For example, infrared and visible energy from the Sun impinge on Earth at an
intensity of 1300 W/m 2 just above the atmosphere. There are other intensity-related units in use, too. The most common is the decibel. For
−3
example, a 90 decibel sound level corresponds to an intensity of 10
W/m 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 W/m 2 .
(a) Calculate the amount of energy that falls on a solar collector having an area of
0.500 m 2 in 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 = Et , the definition of intensity can be written as I = P = E / t , and this equation can be solved
A
A
for E with the given information.
Solution a
1. Begin with the equation that states the definition of intensity:
2. Replace
3. Solve for
I = P.
A
(16.76)
I = E / t.
A
(16.77)
E = IAt.
(16.78)
E = ⎛⎝700 W/m 2⎞⎠⎛⎝0.500 m 2⎞⎠⎡⎣(4.00 h)(3600 s/h)⎤⎦.
(16.79)
5.04×10 6 J,
(16.80)
P with its equivalent E / t :
E:
4. Substitute known values into the equation:
5. Calculate to find
E and convert units:
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
1. Take the ratio of intensities, which yields:
I′ = P′ / A′ = A ⎛The powers cancel because P′ = P⎞.
⎠
I
P/A
A′ ⎝
2. Identify the knowns:
3. Substitute known quantities:
4. Calculate to find
A = 200A′,
I′ = 200.
I
(16.81)
(16.82)
(16.83)
I′ = 200I = 200⎛⎝700 W/m 2⎞⎠.
(16.84)
I′ = 1.40×10 5 W/m 2.
(16.85)
I′ :
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