Question

Loudspeaker manufacturer advertises that their model no. 801 speaker produces a sound pressure level of 87 decibels (db) when a reference test tone is applied. A competing speaker company advertises that their model X-1 speaker produces a sound pressure level of 25 db when fed the same test signal. What is the ratio of the two sound intensities produced by these speakers? If you wanted to find a speaker which produces a sound intensity thrice that of no.801 when fed the test signal, what is its sound pressure

Answer 1

The formula for sound pressure of a noise is defined by
B = (10)(log (I/Io))
where I is the intensity of sound in watts per meters squared and Io is hearing threshold for a human,
lowest sound human can hear: 10^(-12) watts per meters squared

Answer 2

what do I do?

Answer 3

Let me just rewrite everything here so that you can see everything at once:
- We have \(B_{801} = 87\textrm{ dB}\) and \(B_{\textrm{X}-1} = 25\textrm{ dB}\);
- We know that \(B = 10 \log_{10}\left(\dfrac{I}{I_0}\right)\), with \(I_0 = 10^{-12}\textrm{ W m}^{-2}\);
- We want to compute \(\dfrac{I_{801}}{I_{\textrm{X}-1}}\).

You can start by expressing \(I\) in terms of \(B\) to compute \(I_{801}\) and \(I_{\textrm{X}-1}\) first and then take their ratio.

If you notice that \(B\) is defined in terms of \(\log_{10}\), you can also take the difference \(B_{801} - B_{\textrm{X}-1}\) and use the properties of the logarithm to extract the ratio directly.

Can you see how? (:

Answer 4

Nice u stupe

Answer 5

is the answer to What is the ratio of the two sound intensities produced by these speakers? 10^6.2?

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight
is the answer to What is the ratio of the two sound intensities produced by these speakers? 10^6.2?
\(\color{#0cbb34}{\text{End of Quote}}\)
That is correct: \(\dfrac{I_{801}}{I_{\textrm{X}-1}} = 10^{\frac{B_{801} - B_{\textrm{X}-1}}{10}} = 10^{6.2} \quad \ddot{\smile}\)

Can you now answer the next one?

Answer 7

oh right

Answer 8

give me a couple min

Answer 9

is 3 times the sound of intensity of no. 801 (which is 10^-3.3) 10^-9.9 or 10^-1.1?

Answer 10

or 3(10^-3.3)

Answer 11

The sound intensity of 801 is given by:
\(I_{801} = I_0 10^{B_{801}/10} = 10^{-12} \times 10^{87/10} \textrm{ W m}^{-2} = 10^{-3.3}\textrm{ W m}^{-2}.\)
Therefore, thrice this sound intensity is:
\(I = 3 \times I_{801} = 3 \times 10^{-3.3}\textrm{ W m}^{-2}.\)

Answer 12

k thats what I thought

Answer 13

about 9.177

Answer 14

right?

Answer 15

What is 9.177 supposed to be? Notice that the 'sound pressure' of the 801 speaker is 87 dB and the new speaker is supposed to have a larger intensity.