Physics of Sound and Music I week 5
Sound Pressure, power + loudness
In general, if you increase the intensity (ses şiddeti, yoğunluğu) of sound by a factor of 10 (ten times MORE energy
hitting your ear every second), you perceive that to be a little louder. Every factor of 10 increase in intensity
seems like the same "amount louder". It doesn't seem 10 times louder, it seems "one more step" louder.
We have set up a scale that will match your perception better than the actual intensity, and that scale is
called deciBels. (abbreviate dB) (the "bel" is named for Alexander Graham Bell)
as the intensity goes down by FACTORS of 10, the decibels just subtract 10. If intensity goes up by 10 times, decibels
add 10.
the energy flow is MULTIPLYING, but the perception (the decibels) is ADDING.
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
We (arbitrarily) call that sound "zero decibels" or 0 dB. It's near total silence.
(So decibels is still a physical measure, but it relates more personally to how "loud" the sound
"seems". The softest detectable possible sound, I0, will therefore be called "zero dB".)
If you increase the intensity by ten times, so there's 10 times more energy hitting your ear each sec,
We say the new sound is "10 decibels". So MULTIPLYING energy by 10 ADDS 10 to the decibels.
if you have 10*I0 of intensity, you hear a 10 dB sound.
lets get 10 times louder than this. So now we're at 10*10*I0, or 100 times the energy of that softest
sound. But that merely ADDS another 10 dB. So 20 dB corresponds to 100 I0.
Let's go up another factor of 10 in energy. We're now at 10*10*10*I0 = 1000 I0. One THOUSAND
times the energy, but it only added another 10 dB. We're up to 30 dB now. That's what you hear in the
lecture hall before class when no-one is talking, and there's no music on. A quiet room. (Noise still
comes, from e.g. fans, rustling clothes...)
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
The energy rises up 10 TIMES more, but the dB scale only slides up a steady 10 dB.
The intensity is rising by powers of 10, but the dB is increasing LINEARLY (it's adding 10)
That's how your ears work, that's how it FEELS to you.
It's how nature allows us to hear sounds of radically different intensity.
If you increase the intensity by ten times, so there's 10 times more energy hitting your ear each sec.
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
decibel
a way to express the ratio of two quantities, especially very large ( or small ) ratios.
ex
power gain of a stereo amp
= 30dB
it is relative power of the output to the input
so, what does 30dB mean?
if input power to the amp is = 0.01 Watts
and output power of amp is = 10 Watts then,
L₁ = 0.01w and L₂ = 10w
L₂
gain in dB = 10 log –––
L₁
10
= 10 log –––––
0.01
10¹
= 10 log –––––
10⁻²
= 10 log10³
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
log 10
log 10²
log 10³
log 10⁴
...
=1
=2
=3
=4
gain = 10 log10³ = 10x3 = 30dB
ex
we want to double the gain of a stereo signal, what will be the gain in dB?
L₁ = y double that L₂ = 2y
L₂
change in gain dB = 10 log –––
L₁
2y
= 10 log ––– = 10 log 2 = 10 x (0.3) = 3dB
y
10 times the gain = 10 log 10 = 10dB
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
4 times the gain = 10 log 4 = 10dB
∆L = 10 log (2x2)
= 10 log 2²
log Aⁿ = n log A
∆L = 10 x 2 log 2 = 10 x 2 x 0.3 = 6dB
power gain of 5
10
∆L = 10 log (5) = 10 log –––– = 10 [ log 10 - log 2] = 10 [ 1 - 0.3] = 7dB
2
power gain of 400
∆L = 10 log (400) = 10 log ( 10 x 10 x 2 x 2) = 10 [ log10 + log10 + log2 + log2]
= 10 [ 1+1+0.3+0.3] =26dB
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
intensity (ses şiddeti, yoğunluğu)
power
sound intensity ( measurement of energy) = ––––––––––
unit area
=
w
–––––
m²
power = energy/sec
I = intensity
The metric unit of power (which is energy/time) = Watts = w
1 W = 1 Watt = 1 Joule passing by (or transferred) each second.
Sound INTENSITY measures how much energy passes a given area in a given amount of time.
threshold of hearing
Io = 10⁻¹² w/m²
Lı = sound intensity in dB relative to the threshold of sound
I
Lı = 10 log ( ––––– )
Io
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
relationship between intensity w/m² (ses şiddeti, yoğunluğu) and pressure (basınç) N/m² = Pa = Pascals
p²
I = intensity = ––––––––
ρv
ρ = density of air = 1.29 kg/m³
kg
m
kg
ρv = –––––– x ––––– = –––––––––
m³
s
s m²
unit analysis of
p²
N
N
s m²
kg m
kg m
kg
–––––––– = ——— —— ——–––– ––––––– –––––––– = –––––––––
ρv
m²
m²
kg
N s²
N s²
s³
w
kg
intensity = –––––––– = ––––––––––
m²
s³
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
at the threshold of hearing the pressure is po = 2 X 10⁻⁵ N/m² = 2 X 10⁻⁵ Pa = Pascal
express sound intensity in terms of pressure
p
Lp = 10 log ( –––––– ) ² =
2 X 10⁻⁵
Lp = 10 2 log
Lp = 20 log
(
(
p
–––––– )
2 X 10⁻⁵
p
––––––
po
)
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
examples of various sound levels in dB
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
protect your ears
That energy has to go somewhere - ultimately, it becomes heat, and can also cause physical damage to cells. Very loud
sounds are putting energy into your ear, which can cause permanent damage(!)
remember that energy is just power and time, so the longer you listen to that loud music, the more energy is going into
your ear, and the more damage is being done.
There are at least TWO obvious ways to harm your ears then: more intensity, or simply more time... Either way, you end
up with more energy going in to do work on the cells!
According to the text, 100 dB can start to damage the ear. There has been a lot of interest in this
recently as lots of people listen to high volume sound for a lot of time.
If you play in a band or work in a loud place, you really need to learn more about this, and protect your ears
appropriately.
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
ex
if the sound pressure level is 10⁻³ N /m² = 10⁻³ Pa, what is the sound pressure level in dB?
po = 2 X 10⁻⁵ N/m²
Lp = 20 log
Lp = 20 log
(
p
––––––
po
(
10⁻³ N /m²
––––––––––––––
2 X 10⁻⁵N /m²
)
) = 20 log 50 = 20 log (5 x 10) = 20 [ log (5) + log (10)]
sound pressure level is = 20 (0.699+1) = 34dB
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
ex
if the diameter of the eardrum d =7mm, how much force is extended on the eardrum if the
sound pressure level = 120 dB?
force = pressure x area
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
ex
if the diameter of the eardrum d =7mm, how much force is extended on the eardrum if the
sound pressure level = 120 dB?
force = pressure x area
area of the eardrum A= πr²
= 3.14 x 0.007² = 3.85 x 10 ⁻⁵ m²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
ex
if the diameter of the eardrum d =7mm, how much force is extended on the eardrum if the
sound pressure level = 120 dB?
force = pressure x area
area of the eardrum A= πr²
SPL = 20 log
120 = 20 log
6 = log
(
(
p
––––––
po
)
(
p
––––––
po
)
p
––––––
po
)
= 3.14 x 0.007² = 3.85 x 10 ⁻⁵ m²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
ex
if the diameter of the eardrum d =7mm, how much force is extended on the eardrum if the
sound pressure level = 120 dB?
force = pressure x area
area of the eardrum A= πr²
6 = log
(
p
––––––
po
)
= 3.14 x 0.007² = 3.85 x 10 ⁻⁵ m²
in order to solve this, we raise each side of the equation to 10 power
log (p/p0)
p
10⁶ = 10
= ––––––––
po
po = 2 X 10⁻⁵ N/m²
10⁶ x po = p
= 10⁶ x 2 X 10⁻⁵ = 20 N/m²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
ex
if the diameter of the eardrum d =7mm, how much force is extended on the eardrum if the
sound pressure level = 120 dB?
force = pressure x area
area of the eardrum A= πr²
= 3.14 x 0.007² = 3.85 x 10 ⁻⁵ m²
p = 20 N/m²
force = 20 N/m² x 3.85 x 10 ⁻⁵ m² = 7.7 x 10⁻⁴ N
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
3 violins each playing the same note, play at the same sound pressure level (SPL) = 50dB
what will be the resulting SPL be in dB?
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
3 violins each playing the same note, play at the same sound pressure level (SPL) = 50dB
what will be the resulting SPL be in dB? Io = 10⁻¹² w/m²
I₁ + I₂ + I₃
Lp = 10 log ( ––––––––––– )
Io
I
Lp = 10 log ( ––– ) = 50dB
Io
if I₁ = I₂ = I₃ = I
3I
Lp= 10 log ( ––––– )
Io
I
Lp= 10 [ log ( ––––– ) + log 3 ]
Io
=
I
Lp= 10 log ( ––––– ) + 10 log 3
Io
10 log 3 = 4.8
= 50 + 4.8 = 54.8dB
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if violin 1 is playing at 50dB, and violin 2 is playing at 53dB, what will be the resulting SPL in dB?
I₁
violin 1 Lp = 10 log ( ––– ) = 50dB
Io
I₁
10 log ( ––– ) = 50dB
Io
I₁
5 dB = log ( ––– )
Io
log (I₁/Io)
I₁
10⁵ = 10
= ––––
Io
Io = 10⁻¹² w/m²
10⁵ X 10⁻¹² w/m² = I₁
I₁= 10⁻⁷ w/m²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if violin 1 is playing at 50dB, and violin 2 is playing at 53dB, what will be the resulting SPL in dB?
I₂
violin 2 Lp = 10 log ( ––– ) = 53dB
Io
Io = 10⁻¹² w/m²
10⁵·³ X 10⁻¹² w/m² = I ₂
I₂
10 log ( ––– ) = 53dB
Io
I ₂ = 2 X 10⁻⁷ w/m²
total intensity I = I₁ + I₂
I₂
5.3 dB = log ( ––– )
Io
I = 10⁻⁷ w/m² + 2 X 10⁻⁷ w/m² = 3 X 10⁻⁷ w/m²
resulting SPL in dB
log (I₂/Io)
10⁵·³ = 10
I₂
= ––––
Io
I
Lı = 10 log ( –– ) =
Io
3 X 10⁻⁷
Lı = 10 log ( –––––––––– ) =
10⁻¹²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if violin 1 is playing at 50dB, and violin 2 is playing at 53dB, what will be the resulting SPL in dB?
3 X 10⁻⁷
Lı = 10 log ( –––––––––– ) =
10⁻¹²
Lı = 10 log ( 3 X 10 ⁵ ) =
Lı = 10 [ log 3 + log 10 ⁵ ] =
Lı = 10 [ 0.48 + 5 ] = 54.8dB
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
f = 1000 Hz we output from speaker a single frequency at 0dB = 10⁻¹² w/m² this is the threshold of hearing
if we lower down the frequency to 20 Hz, we can no longer hear the sinusoid tone. unless we raise the SPL to the
point that we can hear the tone. e.g. for 50Hz, we need to raise the SPL to 40dB
this graph is mapping out the efficiency of the ear as a function of frequency at different SPL’s.
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
f = 3000 Hz the person indicates that SPl is well above the threshold
the ear is actually most efficient at 3000Hz.
the external ear canal
<--
L
-->
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
calculate the L of a closed pipe with the resonant frequency as 3500hz
L=
λ₁
––––––
4
λ₁ = 4L
f₁ =
v
v
–––––– = ––––––
λ₁
4L
2.43 cm the external ear canal
<--
L
-->
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if 2 violins play at 45dB each, and additional 2 violins play at 55dB each,
what will be the resulting SPL in w/m²?
what will be the resulting SPL in dB?
Io = 10⁻¹² w/m²
I₁
Lı = 10 log ( –––– ) =
Io
I₁
45dB= 10 log ( –––– ) =
Io
I₁
4.5dB= log ( –––– ) =
Io
log (I₁/Io)
10⁴·⁵ = 10
I₁
I₁
= –––––––– = ––––––––
Io
10⁻¹²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if 2 violins play at 45dB each, and additional 2 violins play at 55dB each,
what will be the resulting SPL in w/m²?
what will be the resulting SPL in dB?
Io = 10⁻¹² w/m²
log (I₁/Io)
10⁴·⁵ = 10
I₁
I₁
= –––––––– = ––––––––
Io
10⁻¹²
I₁ = 10⁴·⁵ x 10⁻¹²
I₁ = 10⁴ x 10⁰·⁵ x 10⁻¹²
10⁰·⁵ =
√10 = 3.16
I₁ = 3.16 x 10⁻⁸ w/m²
second violin is also at 45dB so I₂ = 3.16 x 10⁻⁸ w/m²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if 2 violins play at 45dB each, and additional 2 violins play at 55dB each,
what will be the resulting SPL in w/m²?
what will be the resulting SPL in dB?
Io = 10⁻¹² w/m²
3rd and 4th violin are each at 552db
I₃
Lı = 10 log ( –––– ) =
Io
I₃
55dB= 10 log ( –––– ) =
Io
I₃
5.5dB= log ( –––– ) =
Io
log (I₃/Io)
10⁵·⁵ = 10
I₁
I₁
= –––––––– = ––––––––
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if 2 violins play at 45dB each, and additional 2 violins play at 55dB each,
what will be the resulting SPL in w/m²?
what will be the resulting SPL in dB?
Io = 10⁻¹² w/m²
log (I₁/Io)
10⁵·⁵ = 10
I₁
I₁
= –––––––– = ––––––––
Io
10⁻¹²
I₃ = 10⁵·⁵ x 10⁻¹²
I₃ = 10 ⁵ x 10⁰·⁵ x 10⁻¹²
10⁰·⁵ =
√10 = 3.16
I₃ = 3.16 x 10⁻⁷ w/m²
second violin is also at 55dB so I₄ = 3.16 x 10⁻⁷ w/m²
Physics of Sound and Music I week 5
Sound Pressure, power + loudness
multiple source of sound
ex
if 2 violins play at 45dB each, and additional 2 violins play at 55dB each,
what will be the resulting SPL in w/m²?
what will be the resulting SPL in dB?
Io = 10⁻¹² w/m²
total intensity
I₁ + I₂ + I₃ + I₄
3.16 x 10⁻⁸ w/m² + 3.16 x 10⁻⁸ w/m² +3.16 x 10⁻⁷ w/m² + 3.16 x 10⁻⁷ w/m²
0.316 x 10⁻⁷ w/m² + 0.316 x 10⁻⁷ w/m² +3.16 x 10⁻⁷ w/m² + 3.16 x 10⁻⁷ w/m²
I = 6.952 x 10⁻⁷ w/m²
I
Lı = 10 log ( ––––– )
Io
6.952 x 10⁻⁷ w/m²
Lı = 10 log ( ––––––––––––––––– ) = 10 x [ log (7) + log 10⁵] = 10 x 0.84 x 5 = 42dB
10⁻¹² w/m²