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Power and
Intensity
 -Intensity, “I”, of a wave is the power delivered per unit area and is defined by I= P/A,
where P is the rate at which the wave delivers energy (“W”) per unit area (“m2”).
 -Intensity can also be defined as I=(1/2)pv(w)2S2
m. Where p = mass density (in kg/m3), v =
wave speed, w = angular frequency, and Sm = amplitude of sound wave.
 When measuring sound levels, it is common to use the unit for logarithmic function
“decibel” (dB).
 Since logarithmic scales are relative scales, we choose a reference intensity level to be
1000 Hz (about 10 W/m2). Therefore 0 dB corresponds to an intensity I0 = 10-12 W/m2.
 To express any other sound intensity level (B = beta) is “B(I) = 0 dB + 10 log10 (I/I0).
Introduction to sound Intensity and
Power
Typical sound intensity levels
for some common sources
 Sample Question:
 Suppose you are being held captive in a
cell, being tortured for information by
listening to Justin Bieber songs with a sound
intensity of 3.24 W/m2. What is the sound
intensity in decibels (dB)? (Recall B(I) = 0 dB
+ 10 log10 (I/I0))
Solution:
 Recall that I0 = 10-12 and 0 dB = 10 W/m2
 B(I) = 0 dB + 10 log10 (I/I0)
 Plugging in I = 3.24 W/m2 and I0 = 10-12 W/m2 , we get
B(I) = 10 W/m2 + 10 log10 (3.24 W/m2 / 10-12 W/m2) = 125 dB
 If we think about distance and sound intensity, it makes sense that as we move further
away from the source, sound intensity decreases.
 Suppose, then, at some distance (m) we enclose a sphere around the source with
radius (r) = distance (m), then
P = I(4 π r2) *recall I = P/A  P = IA
Here we can see that (4 π r2) is the A (area) of a sphere.
 The equation above states that the power radiated by the source must pass through
the surface of the sphere. This is called Isotropic, which means uniformly in all directions.
 Sample Problem #2
 Suppose you want to escape the inside of a troll’s spherical stomach with a diameter of 12
m by blasting Justin Bieber songs until he throws up. What is the amplitude of the sound
wave given the period of the wave is 0.4 m, the air density in his tummy is 1.20 kg/m3, the
power of the wave is 1.14 kW and the speed of the wave is 343 m/s?
Solution
 Recall (1) P = I(4 π r2) and (2) I=(1/2)pv(w)2S2
m Where “Sm” is the
sound wave amplitude, “p” = density, “v” = wave speed, and “w” =
angular frequency.
 Using equation (1) we can solve for “I” and use it in equation (2) to
solve for amplitude (Sm)
(1) P = I(4πr2)
 (1) We know that P = 1.14 kW = 1,140 W and that radius (r) = ½
diameter (12m) = 6m.
 P = I(4πr2)  1,140W = I(4 π (6.0m)2)  1,140/(4π(6.0m)2) = I
= 2.52 W/m2
(2) I=(1/2)pv(w)2S2
m
 “w” = frequency = 2π /T (“T” = Period), = 2π /0.4 = 5π.
 “v” = velocity = 343 m/s
 “p” = density = 1.2 kg/m3
 “I” = intensity = 2.52 W/m2
 Isolating for Sm we get sqrt [(2I)/(pv(w2))] = Sm
Plugging in the values we get
sqrt [(2*2.52 W/m2)/(1.2 kg/m3 *343 m/s*(5π)2] = 0.00704 m = Sm
Thanks for watching!

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Power and intensity

  • 2.  -Intensity, “I”, of a wave is the power delivered per unit area and is defined by I= P/A, where P is the rate at which the wave delivers energy (“W”) per unit area (“m2”).  -Intensity can also be defined as I=(1/2)pv(w)2S2 m. Where p = mass density (in kg/m3), v = wave speed, w = angular frequency, and Sm = amplitude of sound wave.  When measuring sound levels, it is common to use the unit for logarithmic function “decibel” (dB).  Since logarithmic scales are relative scales, we choose a reference intensity level to be 1000 Hz (about 10 W/m2). Therefore 0 dB corresponds to an intensity I0 = 10-12 W/m2.  To express any other sound intensity level (B = beta) is “B(I) = 0 dB + 10 log10 (I/I0). Introduction to sound Intensity and Power
  • 3. Typical sound intensity levels for some common sources
  • 4.  Sample Question:  Suppose you are being held captive in a cell, being tortured for information by listening to Justin Bieber songs with a sound intensity of 3.24 W/m2. What is the sound intensity in decibels (dB)? (Recall B(I) = 0 dB + 10 log10 (I/I0))
  • 5. Solution:  Recall that I0 = 10-12 and 0 dB = 10 W/m2  B(I) = 0 dB + 10 log10 (I/I0)  Plugging in I = 3.24 W/m2 and I0 = 10-12 W/m2 , we get B(I) = 10 W/m2 + 10 log10 (3.24 W/m2 / 10-12 W/m2) = 125 dB
  • 6.  If we think about distance and sound intensity, it makes sense that as we move further away from the source, sound intensity decreases.  Suppose, then, at some distance (m) we enclose a sphere around the source with radius (r) = distance (m), then P = I(4 π r2) *recall I = P/A  P = IA Here we can see that (4 π r2) is the A (area) of a sphere.  The equation above states that the power radiated by the source must pass through the surface of the sphere. This is called Isotropic, which means uniformly in all directions.
  • 7.  Sample Problem #2  Suppose you want to escape the inside of a troll’s spherical stomach with a diameter of 12 m by blasting Justin Bieber songs until he throws up. What is the amplitude of the sound wave given the period of the wave is 0.4 m, the air density in his tummy is 1.20 kg/m3, the power of the wave is 1.14 kW and the speed of the wave is 343 m/s?
  • 8. Solution  Recall (1) P = I(4 π r2) and (2) I=(1/2)pv(w)2S2 m Where “Sm” is the sound wave amplitude, “p” = density, “v” = wave speed, and “w” = angular frequency.  Using equation (1) we can solve for “I” and use it in equation (2) to solve for amplitude (Sm)
  • 9. (1) P = I(4πr2)  (1) We know that P = 1.14 kW = 1,140 W and that radius (r) = ½ diameter (12m) = 6m.  P = I(4πr2)  1,140W = I(4 π (6.0m)2)  1,140/(4π(6.0m)2) = I = 2.52 W/m2
  • 10. (2) I=(1/2)pv(w)2S2 m  “w” = frequency = 2π /T (“T” = Period), = 2π /0.4 = 5π.  “v” = velocity = 343 m/s  “p” = density = 1.2 kg/m3  “I” = intensity = 2.52 W/m2  Isolating for Sm we get sqrt [(2I)/(pv(w2))] = Sm Plugging in the values we get sqrt [(2*2.52 W/m2)/(1.2 kg/m3 *343 m/s*(5π)2] = 0.00704 m = Sm