Practice: A source emits a sound in the shape of a cone, as shown in the figure below. If you measure the intensity to be 100 W/m^{2} at a distance of 0.5 m, what is the power of the source?

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Concept #1: Sound Intensity

**Transcript**

Hey guys, in this video we're going to talk about sound intensity, intensity is a measure of sound waves that's very that is related to the energy carried by that sound wave but is a more useful measurement to know, let's get to it. Now intensity is the amount of power carried by a wave over the surface area that that wave happens to be spread across, power is fundamental to that wave meaning a source emits a wave at some power and as long as that power isn't dissipated it's going to remain constant as a wave travels through space. For instance in this recording studio where I'm making this video we have sound insulation to protect from outside noises that means whatever sound is passing through the walls with insulation is that insulation is supposed to absorb all the energy in that sound absorb all that power so that sound doesn't pass through into the room that I'm filming in. Now imagine a sound source some source of sound is emitting sound in all directions I draw a figure here sometimes we call this emitting spherically or if you want to get fancy we would say emitting it isotropically. Isoo meaning the same tropically I'm assuming it has something to do with direction but I'm not sure so I'm not going to lie to you guys I just know isotropic means the same in all directions the surface of an area depends upon R squared right the surface area sorry the surface area of a sphere depends on R squared the surface area of a sphere is just 4 pi R squared that means the surface area depends on R square and so the intensity depends on R squared for that very reason that I said assuming there's nothing to absorb the sound then that power is a constant so the only thing that changes is the sound gets further and further out is the radius of the sphere that the sound is spread across, so the intensity changes the further out you go. Two intensities at two different distances are related by the following equation we have I1 in the numerator and I2 in the denominator the way to remember this is it's always the reciprocal, so R2 over R1 squared. If I2 is in the denominator R2 is in the numerator, obviously if you reciprocate both sides the equation still holds.

Let's do a quick example, a speaker emits a sound that you measure to have the intensity of 100 watts per meter squared. When you were 5 meters away from it what would the intensity measured be if you walked 3 meters towards the speaker? So we have R1 is 5 meters and intensity 1 is a 100 watts per meters squared that's just the intensity measured at R1 and we have R2. We were 5 meters away we walk 3 meters towards the source so now we are 2 meters away and I2 is our unknown so I'm going to write I2 over I1 and remember it's the reciprocals so this is R1 over R2 squared and that means I2 is our 1 over R2 squared times I1 which is 5 over 2 squared times a 100 watts per meter squared which is about 625 watts per meter squared so what does that tell us that the intensity increased the closer you got and this is something that you would expect because the closer you got that sphere that contained all that power emitted by the source is smaller so the power density is larger. The intensity is also related to the maximum pressure in a sound wave remember that a sound is just oscillating pressure so it's going to have a maximum pressure and it's going to have a minimum pressure, the maximum pressure is one sorry the intensity equals one half times the maximum pressure divided by the density of a gas that the sound is propagating in times the speed of the sound. So we can do another quick example here air has a density of 1.22 kilogram's per cubic metre if a sound wave has an intensity of 1 times 10 to the 7 watts per square meter what is the maximum pressure of the wave if the air temperature is 0 degrees Celsius? Well what changes with an air temperature of 0 degrees? The speed what changes with the air temperature? The speed of the sound but the intensity, so that's a little bit of foreshadowing there that we're going to have to calculate the speed is P max divided by rho V, now just as a heads up we are dealing with pressure and power here both of which are given by capital P know by context and by your familiarity with these equations that this P here is pressure this is maximum pressure this is not maximum power right. The power doesn't change as the sound travels so rearranging this equation P max is 2I times rho times V the speed of the sound so what is the speed of sound at 0 degrees it's 331 meters per second times 1 plus the temperature in Celsius which is 0 over 273 right and that is just 331 meters per second. So the maximum power is 2 times that intensity 1 times 10 to the 7 times the density 1.22 that density is given right here already in SI units times the speed which is 331 meters per second squared and that is 8.1 times 10 to the 9 Pascals which is a huge huge huge pressure almost 100,000 times atmosphere pressure that would kill you definitely. Alright guys that wraps up this introduction into sound intensity. Thanks for watching.

Practice: A source emits a sound in the shape of a cone, as shown in the figure below. If you measure the intensity to be 100 W/m^{2} at a distance of 0.5 m, what is the power of the source?

Concept #2: Volume And Intensity Level

**Transcript**

Hey guys, in this video we're going to talk about something called intensity level and how it relates to volume. let's get to it. Volume of the sound is not actually a measure of the intensity of the sound wave but it's actually a measurable related quality called the intensity level so when I gave the introduction I said we're talking about the intensity level and how it relates to volume. Well intensity level is volume that's how they are related to one another, really what we want to talk about is how intensity level and thus volume are related to intensity. Now the reason why we use something else than intensity for volume is because volume doesn't really change that much with small changes in intensity there are only noticeably large changes in volume with incredibly large changes in intensity and thus we use a logarithmic scale for intensity which we call intensity level the intensity level which is given by the Greek letter beta is 10 D B times the logarithm of I over I not where D B is our unit the decibel and I am not is called the threshold of hearing and it's 1 times 10 to the -12 watts per meter squared that's the quietest sound the human ear can hear.

Now because this is a logarithm it only changes with factors of 10 so when I becomes a factor of 10 larger logarithm becomes 1 larger or 10 larger because sorry or the intensity level becomes 10 larger because it has the factor of 10 out infront. So it's those powers of 10 increases in intensity lead to noticeable changes in volume but if we go from let's say 100 to 102 watts per meter squared that produces no noticeable change in volume if we go from 100 to a 1000 that produces a noticeable change in volume. So we use a logarithmic scale because only those multiples of 10 in intensity really produce noticeable changes in volume. Alright I threw on some common volumes for you guys here the threshold of hearing is by definition 0 decibels this is because beta right which is 10 D B's log of I over I not if we say I not over I not this is 1 and by definition the logarithm of 1 is 0 so the threshold of hearing the quietest sound that you can possibly hear is 0 decibels a quiet room about 40 decibels conversation about 60 decibels probably what I'm speaking at about now inside of this room speakers in a noisy club I'm from Miami so we got lots of music clubs down here 100 decibels or so threshold of pain when it starts to hurt your ears is 130 decibels just to give you a little bit of context a jet aircraft like a 747, 150 feet away is already above 130 decibels at 140 decibels dangerous volumes are above 150 decibels that's when you start to cause serious damage to your ears in the short term not the long term and theoretically there are volumes that can kill you which are usually theorised to be about above 200 decibels but it's obviously never been proven alright. Now the thing is just because the sound is loud enough it has a volume above 0 decibels doesn't mean that a human can hear it humans can't hear every sound above zero decibels because I'm sorry I minimised myself too quickly because there is a range of frequencies that humans can hear at human hearing is considered to be between 20 hertz and 2000 hertz or 20 kilo Hertz but we can still feel the effects of very very loud sounds even out of the range of hearing for us for instance sounds at dangerous volumes at 150 decibels or more even if they are outside the range of hearing we can feel the pressure coming from the sounds that pressure is still a physical form sort of physical thing and that pressure is putting a force on our body even if we can't hear it.

Let's do an example to close this out, sound is measured to be 25 decibels loud at a distance of 10 meters from the source if you walk 4 meters away from the source of that sound what would be the volume of the sound? So beta 1 is 25 decibels and this occurs at a distance R1 of 10 meters, now beta 2 is our unknown and this occurs at a distance of R2, 4 meters further away than R1 or 14 meters. Now the process to solving this problem is a little bit complicated because or at least complicated algebraically because it involves logarithms but this is pretty much the way you're always going to solve these problems they're all usually presented in the same way except instead of maybe walking away they have you walk towards the sound but it's pretty much the same problem the first thing you're going to start with is the relationship between the intensities the idea is that beta 2 is 10 decibels log of I 2 over I 0 you don't know what I 2 is and you can't find I 2 because you don't know what I 1 is but you can express I 2 in terms of I 1. Once you've expressed I 2 in terms of I 1 then that through some algebraic trickery we can get 10 decibels log of I 1 over I not which is by definition beta 1 which we know so we can solve the answer that way. To find the relationship between I 1 and I 2 we just need to use our regular old intensity relationship right it's the reciprocal so it's R1 over R2. So I 2 is going to equal R1 which is 10 over 14 squared I 1 which is about 0.51 I 1. Here's where that algebraic trickery comes into play beta 2 which is 10 decibels log of I 2 over I not I'm going to replace I 2 with the substitution I just found for I 1 this is 10 decibels logarithm of 0.51 I 1 over I not now for a logarithm anytime you have the multiplication of two numbers in this case what I'm multiplying is 0.51 and I 1 over I not. Those are my two multiplications anytime you have the multiplication of an input for a logarithm you can split that into two logarithms that are summing together. So this is equivalent to 10 decibels logarithm of that first multiplication plus 10 decibels logarithm of that second multiplication term and remember this right here is just beta one by definition and we know what beta one is it's 25 decibels. This right here we can plug into a calculator and we find it's a -2.9 decibels. Alright I'm going to get myself just a little bit of room here to finish out this problem, so beta 2 is 25 decibels minus 2.9 decibels which is 22.1 decibels. Alright and that's pretty much how you're going to solve any of these problems they all are pretty much the same and it comes directly from this logarithmic trick that the log of A times B equals the log of A plus the log of B that's that little logarithmic identity that we want to use. If you guys don't remember your logarithmic identity, your logarithmic algebra just for this part specifically in this chapter you should review it just a little because it will come in handy. Alright guys, that wraps up our discussion on sound intensity and volume. Thanks for watching.

Practice: A source emits sound spherically with a power of 2.2 × 10^{4} W. What is the minimum distance away from this sound that would be considered safe (a volume of 150 dB or less)?

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A human can only hear sound above I 0 = 10 -12 W/m2, called the threashold of hearing. If a 100 W speaker emits sound at 500 Hz, what is the furthest distance from the source you could stand and still hear the sound emitted? What would be the furthest distance if the speaker's tune was changed, and now emits a sound at 100 Hz?

Sound is emitted from a 100 W source at a 50 Hz. From a distance of 10 m, what is the maximum pressure produced by the compression of air from the sound waves? Consider the density of air to be 1.225 kg/m3 and the speed of sound to be 340 m/s.

Standing 15 m from a speaker, the speaker has a volume of 40 dB. What's the furthest distance from the speaker you would still be able to hear it from?
A) 150 m
B) 1,500 m
C) 15,000 m
D) 150,000 m

By how many decibels does the intensity level of a sound increase when you triple the intensity of a source of sound?
A) 9.5 dB
B) 4.8 dB
C) 6.0 dB
D) 3.0 dB
E) 12 dB

Standing 1 m from a speaker, you hear a sound of 50 dB. What would the volume of the sound be if you were to walk 9 m away from the speaker?A) 30 dBB) 40 dB C) 60 dBD) 70 dB

A speaker is emitting energy in the form of sound, at a rate of 200 J every second. If you were 15 m from the speaker, and your ear was pointing directly to the speaker (so sound waves enter directly into your ear canal), how much "sound-energy" (energy in the form of sound) enters your ear canal per second? Assume your ear canal is cylindrical with a diameter of 0.7 cm.

The intensity level for busy traffic is β1 = 81 dB. Two people having a loud conversation have an intensity level of β2 = 70 dB. What is the combined intensity level?

The intensity of the waves from a point source at a distance d from the source is I. At what distance from the sources is the intensity equal to 2I?A) d/2B) d/√2C) d/4D) d/8

The sound intensity is 0.0080 W/m2 at a distance of 10 m from an isotropic (same intensity in all directions) point source of sound.(a) What is the power of the source?(b) What is the sound intensity 5.0 m from the source?(c) What is the sound level 10 m from the source? The threshold of human hearing is I0 = 1.0 × 10 -12 W/m2 .

A sound system emits 100 W of power from a speaker. At 10 m away the intensity is I. How far from the source must you be for the intensity to be cut in half?
A. The intensity remains constant
B. 10√2 m
C. 20 m
D. 40 m

By what amount does the intensity level increase when you triple the intensity of a source of sound? (Hint: subtract the old from the new!)
a) 9.5 dB
b) 4.8 dB
c) 6.0 dB
d) 3.0 dB
e) 12 dB

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