Physics Tutorial: Sound Waves. Intensity and Sound Level

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In this Physics tutorial, you will learn:

  • What are sound waves?
  • How do sound waves propagate in matter?
  • How do sound waves vary with the density of medium?
  • What is audible sound? What are the limits of audibility?
  • What are infra and ultra sound? Where they can be used?
  • What is echo?
  • What is sound intensity and what are the factors affecting it?
  • What is sound level and why do we use a logarithmic scale to describe it?
  • What are loudness and pitch and which quantities are they related to?
Waves Learning Material
Tutorial IDTitleTutorialVideo
11.5Sound Waves. Intensity and Sound Level

Introduction to Sound Waves. Intensity and Sound Level

Suppose you are inside a room and somebody, who is outside the room, is speaking loudly. Can you hear his/her voice? What do you think about penetrating ability of sound waves?

Experiments have shown the speed of sound in metals is more than 10 times greater than in air. What conclusion can you draw about the relationship between speed of sound and the density of medium?

Now, suppose you are on the Moon. Can you hear any voice when somebody is speaking? Why?

This tutorial is focused exclusively on sound waves and their properties, given that these waves are a bit specific and therefore, the approach needed for this purpose is different from the other waves.

Things You Already Know About Sound Waves

So far, we have discussed about waves in general and occasionally we have mentioned sound waves to describe them. Thus, in the previous tutorials of this section, you have learned that:

  • Sound is a mechanical wave as it needs a material medium to propagate. No sound waves exist in vacuum.
  • Sound is a longitudinal wave as its amplitude is in the same direction to the wavelength.
  • Sound (as all the other types of waves) propagates at the same speed within the same medium. The equation of sound is the same as the equation of all waves, i.e.
v = λ × f

where λ is the wavelength and f is the sound frequency.

  • Other equivalent equations we can use for sound waves are:
v = λ/T

where T is the period of sound waves;

v = s/t

where s = N × λ is the total distance travelled by the sound, t = N × T is the total time needed for this process and N is the total number of cycles made by the wave during the time t.

  • Sound waves cause compressions and rarefactions in the medium in which they propagate. Thus, in regions containing compression, the medium's particles are denser than usual while in the regions where there is rarefaction, the medium is less dense than in normal conditions as shown in the figure.
Physics Tutorials: This image provides visual information for the physics tutorial Sound Waves. Intensity and Sound Level

We will extend further the discussion about sound waves based on the above information.

Things You Need To Know About Sound Waves Before Entering Into Details

  • Being a longitudinal wave, sound source causes the medium's particles vibrate back and forth. Therefore, denser the material, easier for the sound to propagates as it is easier for medium particles to find other particles to collide with. Look at the figure.
Physics Tutorials: This image provides visual information for the physics tutorial Sound Waves. Intensity and Sound Level

In the first figure, the material is less dense, so it is more difficult for a particle to find other particles to collide with, while in the second figure the material is denser, so particles collide more easily with each other. Hence, sound waves propagate at a higher speed in denser media such as metals, concrete, etc. Now, you can understand why sound cannot propagate in vacuum, i.e. because there are no particles in the medium through which sound can propagate.

  • The typical speeds of sound in some media are shown in the table below.
Physics Tutorials: This image provides visual information for the physics tutorial Sound Waves. Intensity and Sound Level
  • Sound waves can be modelled as transverse waves on a screen to ease their study, i.e. to make visible the distinction between wavelength and amplitude. A typical example of a sound wave modelling is shown in the figure below.
Physics Tutorials: This image provides visual information for the physics tutorial Sound Waves. Intensity and Sound Level

It is easy to see that wavelength has not changed during the sound wave modelling. Compressions are represented by wave crests while rarefactions by troughs.

  • Sound waves can easily penetrate through solid objects as they propagate in a longitudinal way. Yet, the speed of sound changes when it changes medium. We will explain this phenomenon when discussing about a phenomenon called the "refraction of waves".

Limits of Audibility. Audible Sound. Infra and Ultrasound

Humans are able to hear sounds of frequency ranging from 20 Hz to 20 kHz (20 000 Hz). These values are known as the limits of audibility. This means sound waves, which have a frequency within this range, are audible by humans.

Sound waves of a lower frequency than 20 Hz are known as infrasounds. We cannot hear them as their frequency is smaller than our lower limit of audibility. Infrasound are not used in technology as the energy they carry is practically zero.

On the other hand, sound waves with frequency higher than 20 kHz are known as ultrasound. They are widely used in technology such as in ultrasound screening (echoes), ultrasound detectors, ultrasound distance meters (when calculating the ocean depth), etc. We cannot hear ultrasound as the eardrum of humans can vibrate up to 20 000 times per second (20 kHz).

Hearing ability of a human decrease by aging. The maximum values can be reached around the age of 10.


Echo is the phenomenon of sound reflection when it encounters a large obstacle. As a result, a listener can hear his voice twice: one when he produces the sound and the other when his voice turns back. This helps us calculate the distance from a cliff, a mountain, a building, etc. Let's consider an example.

Example 1

A girl standing near a wall shouts and then she hears the echo of her voice after 3.2 s. Calculate the distance of the girl from the wall. Take the speed of sound in air equal to 343 m/s.

Physics Tutorials: This image provides visual information for the physics tutorial Sound Waves. Intensity and Sound Level

Solution 1

The distance travelled by the sound is twice the distance of the girl from the wall as the sound turns back at the girl's ear after hitting the wall. Also, the speed of sound is constant because sound moves within the same medium (air) during the entire process. Thus, we apply the formula of uniform motion to calculate the distance s of the girl from the wall, i.e.

2s = v × t
= 343 m/s × 3.2 s
= 1097.6 m

Therefore, the distance s of the girl from the wall is

s = 1097.6 m/2
= 548.8 m

Intensity of Sound Waves

Sound waves propagate in all directions like an enlarging sphere. However, the power of sound decreases when moving away from the source as the virtual sphere becomes larger and as a result, the same sound has to propagate in a larger area. This means less sound per unit area falls on a listener or receiver who obviously has a constant area of his/her ear or sound detector. We saw "the sound has a lower intensity when moving away from the source".

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Thus, by definition, sound intensity is the power of sound in the unit of area. Given that the area involved is that of a sphere, we obtain for the sound intensity:

I = P/A
= P/4π × r2

where r is the distance from the source to the receiver, i.e. the radius of the sphere.

The unit of sound intensity is [W/m2] as power is measured in watts and area in square metres.

Example 2

A sound source emits waves of energy equal to 0.05 J every second. What is the sound intensity 10 m away from the source?

Solution 2

In this problem, we have the following clues:

E = 0.05 J

t = 1 s

r = 10 m

I = ?

First, let's work out the sound power. We have

P = E/t
= 0.05 J/1 s
= 0.05 W

Thus, for the sound intensity at 10 m away from source, we obtain

I = P/A
= P/4π × r2
= 0.05/4 × 3.14 × 102
= 3.98 × 10^(-5) W/m2

As you can see from the result, the intensity of sound at 10 m away from the source is very small as sound waves do not carry much energy and therefore, they have a small power. Furthermore, the large distance contributes in a further decrease in the sound intensity as less sound falls in the unit of area.

The lowest intensity representing the faintest sound a human ear can detect is about 10-12 W/m2. This minimum intensity is denoted by I0 and it is known as the "threshold of hearing". The value obtained in the above example for the intensity of sound (3.98 × 10-5 W/m2) is considered as large for humans hearing, so it may be disturbing for people who are subjected to it.

Sound Level

Sound Intensity does not vary linearly with power because distance is raised in power two in the formula that relates sound intensity and power. Therefore, it is not that easy dealing with it in calculations. Hence, scientists prefer to use another physical quantity known as sound level to describe the strength of sound waves. Sound level, L is measured in decibel (dB) in honour to Alexander Graham Bell and it is related to the sound intensity according a logarithmic scale, i.e.

L(dB) = 10 × log(I/I0)

where I is the intensity of the actual sound and I0 is the threshold of hearing. You may have read in the parameters plate of air conditioners or refrigerators the noise in dB these devices produce during their operation. (Noise level is another quantity measured in decibels as it represents irregular sound waves).

Since intensity of sound is the power per unit area, we can also use the equation

L(dB) = 10 × log(P/P0)

to represent the noise level. We can take the threshold sound power as P0 = 10 - 12 W.

Example 3

Sound power in heavy traffic is 0.001 W. What is the value of noise level in dB caused by the heavy traffic?

Solution 3

We have P = 0.001 W = 10 - 3 W and P0 = 10 - 12 W. Thus,

L(dB) = 10 × log(P/P0 )
= 10 × log(10-3/10-12))
= 10 × log10-3-(-12)
= 10 × log109
= 9 × 10
= 90 dB

(Here the logarithm properties log ab = b × log a and log 10 = 1 are used).

The advantage of using a logarithmic scale for calculating the sound level consists on the small range of sound level values when intensity changes too much. Let's consider an example to illustrate this point.

Example 4

A machinery in a factory produces a sound of intensity equal to 100 dB. What is the sound level produced by 40 such machineries if they are turned on simultaneously?

Solution 4

The procedure to solve this problem is as follows: We must calculate the sound intensity produced by a single machinery first, and then multiply it by 40. Then we must calculate the new sound level in dB. Thus,

L1 = 10 × log(I1/I0)
100 = 10 × log(I1/10-12)
100/10 = log I1 - log10-12
10 = log I1 - (-12) × log10
10 = log I1 + 12 × 1
log I1 = -2
I1 = 10 - 2
= 0.01 W/m2

The sound intensity produced by 40 such machineries therefore is

I40 = 40 × I1
= 40 × 0.01 W/m2
= 0.4 W/m2
= 4 × 10-1 W/m2

Hence, the sound level produced by these 40 machineries when they turn on simultaneously is

L40 = 10 × log(I40/I0)
= 10 × log(4 × 10-1/10-12 )
= 10 × log4 × 1011
= 10 × (log 4 + log1011)
= 10 × (log 4 + 11 × log10)
= 10 × (0.6 + 11)
= 10 × 11.6
= 116 dB

You may notice that for a 40 times increase in sound intensity, the sound level increases only by 16 dB (from 100 dB to 116 dB). Therefore, don't neglect the slight changes in the values of sound level as they represent a very big change in sound intensity.

Below some typical values for sound intensity and sound level are given.

Physics Tutorials: This image provides visual information for the physics tutorial Sound Waves. Intensity and Sound Level

Loudness and Pitch

Each voice is characteristic in itself; it has its own unique features that lets us know who is talking even if we do not see the person. We are able to distinct the voice of a man from that of a woman, the voice of a kid from an adult, etc. For this, two features that are special only for sound waves comes to our help. They are loudness and pitch.

Loudness of sound is a phenomenon that depends on the sound amplitude. It is defined as the property of sound used for differentiating between the loud and faint sound. Thus, a loud voice means a voice with a high amplitude while a quiet voice has a small amplitude. Loudness is simply the sound level measured in decibels discussed in the previous paragraph.

We cannot take amplitude of sound and loudness as the same thing because loudness is directly proportional to the square of amplitude, not just to the amplitude itself. This means if the amplitude of a sound wave doubles, the loudness quadruples.

Loudness depends on the energy of sound received by a unit area of ear in the unit of time, as energy of waves (as discussed in our Physics tutorial on Energy and the Power of Waves) depend on the amplitude. Remember the formula for the energy of waves, which is true for sound waves as well:

E = m × A2 × ω2/2

On the other hand, pitch of sound is defined as the feature of sound used for differentiating between the shrill and flat sound. Pitch depends on the frequency of sound waves. Thus, a shrill voice (a high pitch voice) means it has a high frequency and a flat voice (low pitch) means it has a low frequency. For example, the voice of a woman has a higher pitch than that of a man.

Given that energy of waves depend on the frequency (ω = 2π × f), pitch depends on the energy of sound waves as well, albeit not directly (pitch varies directly with the square of frequency).

Example 5

A sound wave has an amplitude of 2 mm and a frequency of 50 Hz. If the amplitude triples and the frequency halves, what happens with loudness and pitch of this sound wave?

Solution 5

From theory, we know that when amplitude of a sound wave triples, loudness increases by a factor of 9, as loudness is proportional to the square of sound amplitude. Therefore, the new amplitude becomes 2 mm × 9 = 18 mm.

Also, from theory we know that pitch varies directly with the square of frequency. This means when the sound frequency halves, pitch decrease by a factor of 4. Therefore, the new pitch becomes 50 Hz / 4 = 12.5 Hz (it becomes infrasound).

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