Sound Wave Power and

Intensity

PHYS 101

What is Wave Intensity?

• Wave intensity is the average power of a wave as it is travelling through a

space.

• Wave intensity is generally measured with the decibel scale (units: dB).

• The higher a sound waves intensity, the louder the sound will be perceived,

although a wave with an intensity of 0 dB is still making a sound.

Intensity = Power/Area

• Power is the rate at which a sound transfers its energy, measured in joules per

second (j/s) or watts (W).

• In the case of spherical waves, the formula for the area is A = 4πr^2

• The further from the source a sound wave travels, the more the sound wave

will be reduced in intensity.

As a sound wave travels further from its source, its area will increase but its

power will remain constant. Since Intensity = Power/Area, this means that the

Intensity will also decrease.

Poweravg=(1/2)pvω^2*A*smax^2

• This is the formula for finding the average power in a sound wave

• p is the sound pressure, measured in pascals (Pa)

• v is the particle velocity, which is measured in meters per second (m/s)

• ω is the angular velocity which can be calculated by the formula: 2πƒ or 2π/T

• ƒ is the frequency of the wave and T is its period

• A is the spherical area of the wave

Poweravg=(1/2)pvω^2*A*smax^2

• Since the force of the wave is related to the change in pressure caused by the wave, the force can also be calculated as:

• F = p*v*ω*A*smaxsin(kx- ωt)

• p is the sound pressure measured in pascals (Pa).

• v is the particle velocity

• smax is the maximum possible particle displacement from the waves equilibrium position, it can be calculated by multiplying the particle velocity by the change in time

• k is the angular wavenumber which is calculated by the formula k = 2π/λ, with λ being the length of the wave.

Alternate forms of the equation

• Since Poweravg=(1/2)pvω^2*A*smax^2 and Intensity = Power/Area, we can

eliminate area from the equation to create the formula:

• I = (1/2)pv(ωsmax)^2

• This equation can also be written in relation to the Pressure Amplitude since the

Pressure Amplitude is equal to ω*smax which results in the equation:

• I = ((ΔPmax)^2)/(2pv)

Visual examples of a sound waves wavelength,

amplitude, frequency and period.

Problem #1

• A speaker is giving off sound waves in all directions. You are standing 5 meters away from the speakers and experience 160 dB of wave intensity. If you moved back so that you were 20 meters away from the speaker, what would be the intensity of the sound waves when they reached you?

• A. 80 dB

• B. 40 dB

• C. 10 Db

• D. None of the above

Problem #1 - Solutions

• The answer is C, 10 dB

The intensity of a sound wave is proportional to 1/r^2. Since 160dB = 1/r^2

at 5 meters, at 20 meters, the intensity is equal to I = 1/(4r)^2, resulting in the

intensity being 1/16 of the intensity that it was at 5 meters from the speaker.

•≈•≈

Problem #2

• A spherical sound wave has an intensity of 20 dB at 10 a point 5 meters from the source.

• i.) What is the average power of the sound wave?

• ii.) Assuming the wave has a particle velocity of 3.0 m/s, generate 30 pascalsof pressure, and has an angular frequency of π/5 radians per second, find the maximum particle displacement.

• iii.) What effect would increasing the wave frequency of a sound wave have on its Power, area, intensity, particle displacement, and particle velocity?

Problem #2 - Solutions

• i.) What is the average power of the sound wave?

• As mentioned earlier, the I = Power/Area or I = Power/(4πr^2)

• We know I = 20 dB and the radius is 5 meters therefore:

• 20dB = P/(4π*(5m^2) which we can rearrange to find the wave power.

• P = 20dB*(4π*25m), P = 20dB*(100π)

• We find that the average power of the wave is approximately 6283 Watts.

Problem #2 - Solutions

• ii.) Assuming the wave has a particle velocity of 3.0 m/s and generate 30

pascals of pressure, and has an angular frequency of π/5 radians per second,

find the maximum particle displacement.

The values we are given are: v = 3.0m/s, p = 30 Pa, ω = π/5 rads/sec, we also

know the wave intensity is 20 dB. Using this, we can plug these values into the

equation: I = (1/2)pv(ωsmax)^2 to find the value of the maximum particle

displacement.

Problem #2 - Solutions

• 20 dB = (1/2)(50 Pa)*(3.0 m/s)*((π/5 rads/sec)*smax)^2

• By rearranging the values in the equation we can produce the following equation:

• (square root(((20dB*2)/(30Pa*3.0m/s)))/(π/5rads/sec) = smax

• smax = square root(40 dB/(90Pa*m/s))/(π/5rads/sec)

• By inputting these values, we find that the maximum particle displacement is approximately

0.71 meters.

Problem #2 - Solutions

• iii.) What effect would increasing the wave frequency of a sound wave have

on its power, intensity and spherical area?

• Since ω = 2πƒ, changing the frequency only directly affects the angular

frequency of the wave, however, by determining what properties of the wave

are affected by the angular frequency, we can determine the effect a change

in wave frequency would have on them.

Problem #2 - Solutions

• Power: Since Poweravg=(1/2)pvω^2*A*smax^2, an increase in the wave frequency would result in an exponential increase in the wave’s power.

• Spherical Area: Since A = 4πr^2, a change in frequency would not affect the spherical area of the sound wave since angular frequency is not one of the variables that affects it.

• Intensity: Since I = Power/Area, an increase in the wave frequency would also result in an increase in the intensity of the wave since it would cause an increase in the waves power and it was already determined that it would not increase the spherical area of the sound wave.

Sources Cited

Sources of Images of Spherical Sound Waves:

• http://physics.stackexchange.com/questions/130035/the-inverse-square-law-of-sound-through-solids

• http://www.s-cool.co.uk/a-level/physics/progressive-waves/revise-it/progressive-waves

Source equations, definitions, and facts:

• Physics for Scientists and Engineers - An Interactive Approach