intensity level, dbs, logs

[arc5005]

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If the intensity level of one violin is 24 dB, what can be done to achieve an intensity level of 30 dB?

A. play the violin 1.25 times faster.
B. play a violin with strings that are 1.25 times longer.
C. play four violins simultaneously.
D. play sixteen violins simultaneously.

C) play four violins simultaneously.



First, the intensity made by one violin is 24 dB. To increase to 30 dB, the # of violins must increase. Longer strings and greater playing rate will shift the frequnecy and wave speed, not change the intensity. This eliminates choices A and B. Because the intensity value is a # obtained by multiplying 10, to get the intensity to increase by 6 dB, the intensity must increase by a factor that has a log valu e of 0.6. Log of 10 is 1.0, therefore the value must be less than 10. This eliminates D.


ß = 10 (log (I/I0) = 24 therefore, (I/I0) = 10^2.4


This is the intensity of one violin. To get the intensity to 30 dB, we need the following:
(I/I0) - 10^3.0 = 10^2.4 ∝ 10^0.6


Because the intensity must increase by 10^0.6, which equals 4, the number of violins must increase by 4. Note that we have made use of a log identity here, but it is possible to compelte this problem without knowing it, if you remember how logs behave.


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guess i'm a little confused on the log stuff here. 10^0.6 is log 10 ^ 0.6? or how does this work again?

 

[NextStepTutor_1]

Hi @arc5005 -

Personally, I would tackle this a little bit differently. Eliminate A and B because they're irrelevant -- we're trying to increase intensity by playing extra violins simultaneously. Then work backwards for C and D, with the understanding that each 10-point increment on the dB scale corresponds to increasing intensity by a magnitude of 10, so a 10x increase in intensity would correspond to an extra 10 points on the dB scale. We're looking for an increase of 6 dB, which means that the intensity must increase by less than 10x. This rules in choice C and rules out choice D.

Anyway, just my two cents about how to approach this problem, because I don't think that it's super realistic to be able to estimate 10^0.6 = 4 w/out a calculator, and at least to my mind, working backwards is way more conceptually straightforward here and gives you fewer opportunities to get lost in a thicket of math and make a pure math error -- instead, working backwards allows you to ground your answer in an understanding of the principles of how dBs are defined.

 

[Zenabi90]

Hi @arc5005 -

Personally, I would tackle this a little bit differently. Eliminate A and B because they're irrelevant -- we're trying to increase intensity by playing extra violins simultaneously. Then work backwards for C and D, with the understanding that each 10-point increment on the dB scale corresponds to increasing intensity by a magnitude of 10, so a 10x increase in intensity would correspond to an extra 10 points on the dB scale. We're looking for an increase of 6 dB, which means that the intensity must increase by less than 10x. This rules in choice C and rules out choice D.

Anyway, just my two cents about how to approach this problem, because I don't think that it's super realistic to be able to estimate 10^0.6 = 4 w/out a calculator, and at least to my mind, working backwards is way more conceptually straightforward here and gives you fewer opportunities to get lost in a thicket of math and make a pure math error -- instead, working backwards allows you to ground your answer in an understanding of the principles of how dBs are defined.

I second this approach. It saves both time and mental energy, both of which are at a premium during this exam.

If you're not an instrument person, just think about people talking.

Recognize that A and B, talking faster or in a deeper voice will not make you louder to a microphone. How do you make a person louder? Yes, they can talk louder, or we can just have more people.

Recognize that decibel is a log scale, so a 6 db jump is definitely louder, but less than 10x louder.

I would guess that 16 people all trying to talk at the same time is likely to be at least 10x louder than that one person, so 4 people sounds more reasonable. And thus, with a confidence level of like 90%, answer C and move on and if you have way too much time at the end of the section and it's still bugging you, try to come back and justify your answer with the math.




And for the sadists out there (like me) who can't rest without trying the math in your head...

So when you have a decimal log, the easiest way to understand it is to turn it into a fraction. 0.6 is the same as 6/10. Which means 10 ^ 0.6 is equal to 10 ^ (6/10), or (10^ 1/10) ^ 6. In english, 10 to the power of 0.6 is the tenth root of 10, taken to the sixth power.

Recognize that 10^0.6 = 10 ^ (6/10) = (10^3/5). And yes it's very difficult to do a fifth-root in your head. BUT 3/5 is close to 1/2 (0.6 vs 0.5). The square root of 10 is somewhere between 3 (sqrt of 9) and 4 (sprt of 16) so I'd say something like 3.12. But we want 10^0.6, not 0.5, so let's bump it up to the next nearest whole number, which is 4!

Want it closer than rounding up from 3.1 to 4? 10 ^ 6/10 is even closes to 10^2/3. So, 10^1/3 is somewhere between 2 (cube root of 8) and 3 (cube root of 27) and much closer to 2, so let's say 2.1 and then square it, since it's 10^2/3, and 2.1 x 2.1 is 4.41 (which is easily done on scratch paper. But remember, we want 10^0.6, not 0.66, so round it down to the closest integer which is, again, 4!