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Wize University Physics Textbook (Master) > Sound & Hearing

Sound Waves

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Sound Waves





Sound waves are longitudinal mechanical waves.

  • The speed of the sound in each medium is given by:
 v=Bρ \boxed{ \ v=\sqrt{\frac{B}{\rho}} \ }


  • BB is the Bulk modulus in PaPa, and it measures the substance's resistance to uniform compression
  • ρ\rho is the density of the medium

Watch Out!
There needs to be an oscillating medium for there to be sound.


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  • The sound intensity level β\bcth\beta (units are deciBells, denoted dBdB) is defined as:
 β=(10 dB) log⁡II0 \boxed{ \ \beta=(10 \ dB) \ \log\frac{I}{I_0} \ }

  • II is the intensity of a sound wave
  • I0I_0 is the reference intensity chosen to be the threshold of human hearing:

 I0=10−12 W/m2 \boxed{ \ I_0=10^{-12} \ W/m^2 \ }




Exam Tip
The intensity of 1 W/m21 \ W/m^2 is the threshold of pain.




Wize Concept
The frequency of each sound is its pitch.
  • Humans can hear in the range of 20 Hz (low pitches) to 20 kHz (high pitches)
  • Sounds above this are called ultrasonic, and below this, infrasonic.

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Properties of Sound Waves


Sound is a longitudinal pressure wave.
  • The speed of sound in air depends on the temperature.
vw=(331ms)Tk273v_w=\left(331\frac{m}{s}\right)\sqrt{\frac{T_k}{273}}
  • Where Tk is the absolute temperature of air in kelvin (Tk=Tc+273T_k=T_c+273).
  • The speed of sound in air at 0ºC is 331 m/s.

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Sound Waves

  • Sound waves are longitudinal mechanical waves.
  • There needs to be an oscillating medium for there to be sound. So there's no sound in space!
  • Sound is characterized by pitch (frequency) and loudness (intensity).
  • Humans can hear in the range of 20 Hz (low pitches) to 20 kHz (high pitches)
  • Sounds above this are called ultrasonic, and below this, infrasonic.

  • The sound intensity level β\beta of a sound wave is defined as:
β=(10dB)log⁡II0\beta=\left(10dB\right)\log\frac{I}{I_0}

  • deciBels are the units of β\beta and are abbreviated dBdB
  • IIis the intensity of a sound wave
  • Remember: this is proportional to the inverse square of the distance from the source
  • I0I_0is a reference intensity chosen to be 10−12W/m210^{-12}W/m^2 which is the threshold of the human hearing
  • I=1W/m2I=1W/m^2is the threshold of pain

  • There is also psychology and neuroscience involved in what we can hear and how loud we can hear it. Some examples from physics don't always line up with human experience of sound.
  • Cocktail party problem
  • Human sensitivity to different frequencies

Intensity Shortcut Formula

We can write the sound intensity formula β=(10 dB) log⁡II0\beta=(10 \ dB) \ \log\dfrac{I}{I_0} twice: for the same source, but at different distances, so the intensity will be different at each location. Combining the two formulas allows us to compare the two decibel levels without having to know the reference intensity I0I_0:

 β2−β1=10 log⁡I2I1 \boxed{ \ \beta_2-\beta_1=10\ \log\frac{I_2}{I_1} \ }











The equivalent exponential form of this is:
 I2I1=10 0.1(β2−β1) \boxed{ \ \dfrac{I_2}{I_1}=10^{\ 0.1(\beta_2-\beta_1)} \ }



Example: Intensity Shortcut Formula


a) You are at a party close to a speaker and hear a 125125 dB sound. Then you move 1010 m away and only hear 9292 dB. How much more intense was the sound you heard when you were close to the speaker?

b) You stand a certain distance away from a speaker and hear a 100100 dB sound. What is the new decibel level if you have 55 speakers instead, at the same distance?


Part a)


Let's use the shortcut formula with #2 close to the speaker and #1 farther away:

β2=125\beta_2=125 and β1=92\beta_1=92

Then we have:

I2I1=10 0.1(β2−β1)=100.1(125−92)=1995≈2000\dfrac{I_2}{I_1}=10^{\ 0.1(\beta_2-\beta_1)}=10^{0.1(125-92)}=1995\approx2000

This is the same as saying I2=2000 I1I_2=2000 \ I_1 , which means that the sound close to the speaker was 20002000 times more intense than the sound farther away.


Part b)


The equation for just one speaker is:

β=10 log⁡II0=100\bct{\beta=10 \ \log\dfrac{I}{I_0}=100} (dB)

If we use five speakers, the intensity will also increase by a factor of five: the new intensity will be Inew=5II_{new}=5I.

Therefore the new decibel level is:

βnew=10 log⁡InewI0\beta_{new}=10 \ \log\dfrac{I_{new}}{I_0}

=10 log⁡5II0=10 \ \log\dfrac{5I}{I_0}

Separate the insides of the log:

=10 log⁡5 + 10 log⁡II0=10 \ \log5 \ + \ \bct{10 \ \log\dfrac{I}{I_0}}

=10 log⁡5 + β=10 \ \log5 \ + \ \bct{\beta}

=10 log⁡5 + 100=10 \ \log5 \ + \ \bct{100}

=107=107 (dB)

Practice: Sound Wave Intensity

Answer the following two questions:
a) The intensity of a sound wave 55m away from its source is equal to 10−510^{-5}W/m2. How far from the source can one hear its sound?

Example: Sound Under Water


A loudspeaker is broadcasting a single frequency of 440440 Hz. We would like to know the wavelength of this sound when it penetrates under water. If the bulk modulus of water is 2.22.2 GN/m2 and water density is 10001000kg/m3, find the sound wavelength under water.


First, find the speed of sound in water:

v=Bρ=2.2×1091000=1483.2v=\sqrt{\dfrac{B}{\rho}}=\sqrt{\dfrac{2.2\times10^9}{1000}}=1483.2 (m/s)


The frequency doesn't change when the wave passes through a different medium. But the velocity and wavelength change (they are directly proportional):

v=λf   →   λ=vf=1483.2440=3.37v=\lambda f \ \ \ \to \ \ \ \lambda=\dfrac{v}{f}=\dfrac{1483.2}{440}=3.37 (m)

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a) The intensity of a sound wave 5 m5\ maway from its source is equal to 10−5 wattm210^{-5}\ \frac{watt}{m^2}? How far from the source can one hear its sound?

b) Another sound source has the intensity level equal to 65 dB65\ dBat 2 m2\ mfrom the source. What is the intensity of the sound if you double the distance?



a) I(5)=10−5 Wm2I\left(5\right)=10^{-5}\ \frac{W}{m^2}
I1I2=r22r12\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2} minimum of hearing =I2=10−12 Wm2=I_2=10^{-12}\ \frac{W}{m^2}→r22=I1I2r12=25×10−510−12=25×107\to r_2^2=\frac{I_1}{I_2}r_1^2=25\times\frac{10^{-5}}{10^{-12}}=25\times10^7

→r2=15811.4m=15.811 km\to r_2=15811.4m=15.811\ km

b) β=65 dB\beta=65\ dB at r=2 mr=2\ m
β=10log⁡(II0)→65=10log⁡(II0)\beta=10\log\left(\frac{I}{I_0}\right)\to65=10\log\left(\frac{I}{I_0}\right)

→6.5=log⁡(II0)→10+6.5=II0=I10−12→I=10−12×106.5=10−5.5 Wm2\to6.5=\log\left(\frac{I}{I_0}\right)\to10^{+6.5}=\frac{I}{I_0}=\frac{I}{10^{-12}}\to I=10^{-12}\times10^{6.5}=10^{-5.5}\ \frac{W}{m^2}

at r=2r=2
I1I2=r22r12→I2=r12r22I1=10−5.5×(12)2=7.9×10−7 Wm2\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}\to I_2=\frac{r_1^2}{r_2^2}I_1=10^{-5.5}\times\left(\frac{1}{2}\right)^{^2}=7.9\times10^{-7}\ \frac{W}{m^2}