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17.3 Sound intensity

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By the end of this section, you will be able to:
  • Define the term intensity
  • Explain the concept of sound intensity level
  • Describe how the human ear translates sound

In a quiet forest, you can sometimes hear a single leaf fall to the ground. But when a passing motorist has his stereo turned up, you cannot even hear what the person next to you in your car is saying ( [link] ). We are all very familiar with the loudness of sounds and are aware that loudness is related to how energetically the source is vibrating. High noise exposure is hazardous to hearing, which is why it is important for people working in industrial settings to wear ear protection. The relevant physical quantity is sound intensity, a concept that is valid for all sounds whether or not they are in the audible range.

Photograph shows a roadway crowded with cars and motorcycles in Delhi.
Noise on crowded roadways, like this one in Delhi, makes it hard to hear others unless they shout. (credit: “Lingaraj G J”/Flickr)

In Waves , we defined intensity as the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = P A ,

where P is the power through an area A . The SI unit for I is W/m 2 . If we assume that the sound wave is spherical, and that no energy is lost to thermal processes, the energy of the sound wave is spread over a larger area as distance increases, so the intensity decreases. The area of a sphere is A = 4 π r 2 . As the wave spreads out from r 1 to r 2 , the energy also spreads out over a larger area:

P 1 = P 2 I 1 4 π r 1 2 = I 2 4 π r 2 2 ;
I 2 = I 1 ( r 1 r 2 ) 2 .

The intensity decreases as the wave moves out from the source. In an inverse square relationship, such as the intensity, when you double the distance, the intensity decreases to one quarter,

I 2 = I 1 ( r 1 r 2 ) 2 = I 1 ( r 1 2 r 1 ) 2 = 1 4 I 1 .

Generally, when considering the intensity of a sound wave, we take the intensity to be the time-averaged value of the power, denoted by ⟨ P ⟩ , divided by the area,

I = ⟨ P ⟩ A .

The intensity of a sound wave is proportional to the change in the pressure squared and inversely proportional to the density and the speed. Consider a parcel of a medium initially undisturbed and then influenced by a sound wave at time t , as shown in [link] .

Picture is a drawing of a parcel of a medium initially undisturbed and then influenced by a sound wave. A sound wave moves through the medium at time t, and the parcel is displaced and expands in the displacement direction.
An undisturbed parcel of a medium with a volume V = A Δ x shown in blue. A sound wave moves through the medium at time t , and the parcel is displaced and expands, as shown by dotted lines. The change in volume is Δ V = A Δ s = A ( s 2 − s 1 ) , where s 1 is the displacement of the leading edge of the parcel and s 2 is the displacement of the trailing edge of the parcel. In the figure, s 2 > s 1 and the parcel expands, but the parcel can either expand or compress ( s 2 < s 1 ) , depending on which part of the sound wave (compression or rarefaction) is moving through the parcel.

As the sound wave moves through the parcel, the parcel is displaced and may expand or contract. If s 2 > s 1 , the volume has increased and the pressure decreases. If s 2 < s 1 , the volume has decreased and the pressure increases. The change in the volume is

Δ V = A Δ s = A ( s 2 − s 1 ) = A ( s ( x + Δ x , t ) − s ( x , t ) ) .

The fractional change in the volume is the change in volume divided by the original volume:

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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