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About David Montiel
Expertise
PLEASE READ THIS BEFORE POSTING YOUR QUESTION: While I am glad to offer HELP with problems, I refuse to SOLVE problems. There is a difference. Please do not just copy a question from a texbook or assignment. I WILL REJECT IT. I will only accept questions, if in addition to the statement of the problem, you describe in detail what you have tried so far in order to solve it, and what exactly is the problem you are encountering. In other words where/why are you stuck? I can answer general physics questions up to the level of first or second year undergraduate courses. Topics I am familiar with include: classical mechanics (kinematics, dynamics, etc.), electricity and magnetism, optics, thermodynamics, and special relativity. I may not answer questions at an advanced undergraduate or graduate levels. Also please do not expect help with algebra or arithmetic (performing numerical calculations).

Experience
I have been a private physics tutor for about eight years. I also have six years of experience a teaching assistant (for both theory and lab) and occasional substitute lecturer in Physics.

Education/Credentials
B.S. in Chemical Engineering M.S. in Chemical Physics PhD candidate (5th year) in Physics.

You are here: Experts > Science > Physics > Physics > How does sound travel

Physics - How does sound travel

Expert: David Montiel - 11/3/2009

Question
QUESTION: Hi,
My question is how does sound travel through the air? Is it like a radio wave? And is sound inside a car, more bad for your ears then listening to music outside a car.

Thanks for your help
Regards
Jeff

ANSWER: Hello Jeff,

No, sound waves and radio waves have a different nature. Radio waves belong to a type of waves called "electromagnetic". Another example of this is light; in fact, you could say that radio waves are a kind of invisible light... In contrast, sound waves are "mechanical" in nature. Unlike radio waves, sound waves need a medium to travel (like water or air). Light can also travel through water or air but it doesn't have to; it can also travel through empty space.

OK so how do sound waves travel? Let's say you have a speaker. When the speaker emits a sound, it's membrane vibrates (moves back and forth). This vibration makes the air around the speaker to also move back and forth. Then, the air around the speaker also moves the air around it, and so on. So that's how sound travels through space. The vibration of the molecules in the air eventually reaches your ear and then you hear the sound.

Now, referring to the last part of your question, what determines how bad sound can be for your ears is intensity (how loud the sound is). And the only thing that matters is how loud the sound is when is reaches your ears, not where it's coming from or where you are. Very loud sounds can be damaging to your hearing, whether they are coming from headphones, your stereo, or speakers at a rock show, for instance.

There is an interesting thing about the sound inside a car though. And that is that sound can also bounce from the walls (in this case windows) of the car; and this can increase the intensity of sound when you're inside a car with the windows closed. So yes, if for instance, you're listening to music in a convertible, you wouldn't hear it as loud as you would in a normal car (given that you have same "volume" setting).

Hope this is clear.

Cheers,
David

---------- FOLLOW-UP ----------

QUESTION: How are decibels measured? They say 85 and above decibels are bad for your ears.

ANSWER: Decibels (dB) are a measure of sound intensity, but on a logarithmic scale. The formula is:

dB = 10*log( I / I0)

where I is the intensity of the sound where you hear it, and I0 is the minimum sound intensity an average human can hear.

I0 = 1*10^-12 W/m^2

Also the "log" in the equation is the "base 10" logarithm.

What this equation means is that every time intensity gets 10 times bigger, decibels increase by 10. Note that there is a difference. For instance if you hear a sound of 70 dB, and then increase the intensity to be ten times bigger, that gives you a sound of 80dB.

I don't exactly know how you measure decibels but you probably use some kind of microphone to measure how loud the sound is.

Hope this is clear,
David

---------- FOLLOW-UP ----------

QUESTION: Hi again,
Is IO always equal to = 1*10^-12 W/m^2? Also what does W and m stand for?

Also why are logarithms used? I've never heard of them before?

Thanks for all the help
Regards
Jeff

Answer
Hello Jeff,

Yes, you always use I0=1*10^-12 W/m^2 in the formula.

...although of course this number does not always correspond to the real hearing intensity threshold. That varies from person to person.

Also
W = Watts (units of power)
m = meters

Now to your second question...

Logarithms are a convenient whenever you want to express a quantity that varies within a wide range (several orders of magnitude). Imagine, that a variable Q represents some physical quantity. Suppose that whenever Q is measured in different situations, it results in anything form Q=0.00008 to Q=400,000,000. There is a huge range of possible values! In those cases, wouldn't it make sense to express Q in terms of the number of zeroes? We could define a quantity (lets call it P) that measures the order of magnitude of Q.

It turns out that this is exactly what the logarithm does:

log(1)=0,
log(10)=1,
log(100)=2,
...
log(1,000,000)=6,

Do you see what's going on here? The (base 10) logarithm of a number "counts" the number of zeroes of that number. So in our example above we can use

P=log(Q)

We use P instead of Q just because it's more convenient.

Same with sound. Intensity (I) varies so much that it makes more sense to express it in terms of its logarithm (dB).

Finally there is another (even more) important reason to use log(I/I0), and that is that our ears actually perceive sound on a logarithmic scale.

What I mean by this is, suppose we hear a sound (let's call it sound "A") of intensity IA such that IA/I0=100. This corresponds to 20 decibels: [10*log(100)].

Now suppose we hear a second sound "B" of intensity IB such that IB/I0=1000. This corresponds to 30
decibels: [10*log(1000)].

Clearly sound "B" has 10 times more intensity than sound "A". However, it turns out that our ears will not perceive the sound to be 10 times louder. Actually our ears will hear the sound to be proportional to the number of decibels, not intensity. So in this case sound "B" will be 50% louder than sound "A".

Hope this is clear,
David

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