Comparing Pump to Dishwasher Intensity: 10x Difference

[karush]

$d=10\log\left({\frac{{P}_{}}{{P}_{0}}}\right)$
$d$=decibels
$P$ is intensity of the sound
$P_0$ is the weakest sound that the human ear can hear.

If a water pump is $d=50$ and a dishwasher $d=62$
How many times more intense is the pump to dishwasher.

Using exponential form then calculating

$\frac{{P}_{}}{{P}_{0}}=10^{50}$
and
$\frac{{P}_{}}{{P}_{0}}=10^{62}$
Now what?

 

[MarkFL]

I would take:

\(\displaystyle d=10\log\left(\frac{P}{P_0}\right)\)

And solve it for $P$ to get:

\(\displaystyle P=P_010^{\frac{d}{10}}\)

And so we find the ratio $r$ of the intensity of the pump to that of the dishwasher to be:

\(\displaystyle r=\frac{P_010^{\frac{50}{10}}}{P_010^{\frac{62}{10}}}=10^{-1.2}\approx0.063095734448\)

 

[karush]

i postered this on LinkedIn believing it might draw some into the forum it immediately got 14 views

Spoiler: LinkedIn

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