## Cats can scratch your eyes out, but can they meow your ears out?

Recently, I came across a post looking at the question "If all the cats in the world "meowed" at the same time, how loud would the sound be?". The video is below.

Because I was proctoring tests, I had some time, so I decided to go a little further. I started with a basic assumption that cats like to be about 1 meter away from other cats. Starting here, I came across some neat mathematics.

### How big?

Let us assume that the approximately 600 million cats
in the world want to be 1 meter apart. If we set them up on circles with radii of 1 meter, 2
meters, 3 meters, and so on, we can put 6 cats on the first circle, 12 on the
second, 18 on the third, etc. How many circles do we need?

\(6 + 12 + 18 + .............n = 6 \times (1 + 2 + 3 + .......m) = 600,000,000\)

and using the Gaussian sum formula for the first m integers (the legend about how Gauss derived this can be found here)

which gets around 14,000 circles.

### How loud?

Obviously, a cat further away will sound quieter, but there are more of them further away. How do we calculate that balance? As the video illustrates, we can add sound intensity levels, not decibel levels. Let us use the sound intensity level for a cat meow from 1 meter away as \({{I_{{\rm{cat}}}}}\). Remember that sound intensity diminishes as the inverse of the distance squared. So the total sound intensity of the 6 cats in the first circle will be \(6 \times \frac{{{I_{{\rm{cat}}}}}}{{{1^2}}}\), the intensity of the 12 cats in the second circle will be \(12 \times \frac{{{I_{{\rm{cat}}}}}}{{{2^2}}}\). The total intensity for all 14,000 circles will be

\({I_{total}} = 6 \times \frac{{{I_{{\rm{cat}}}}}}{{{1^2}}} + 12 \times \frac{{{I_{{\rm{cat}}}}}}{{{2^2}}} + 18 \times \frac{{{I_{{\rm{cat}}}}}}{{{3^2}}} + ......14,000 \times 6 \times \frac{{{I_{{\rm{cat}}}}}}{{14,{{000}^2}}}\).

A rearrangement gives

\({I_{total}} = 6 \times 1 \times \frac{{{I_{{\rm{cat}}}}}}{{{1^2}}} + 6 \times 2 \times \frac{{{I_{{\rm{cat}}}}}}{{{2^2}}} + 6 \times 3 \times \frac{{{I_{{\rm{cat}}}}}}{{{3^2}}} + ......6 \times 14,000 \times \frac{{{I_{{\rm{cat}}}}}}{{14,{{000}^2}}}\).

Which gives

\({I_{total}} = 6 \times {I_{{\rm{cat}}}} \times \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ....... + \frac{1}{{14,000}}} \right)\).

Notice the harmonic series in the parentheses. Using the definition of the natural logarithm,

\(\int_1^n {\frac{1}{x}} dx \equiv \ln \left( n \right)\) ¹,

we can approximate the sum of this harmonic series as

\({I_{total}} = 6 \times {I_{{\rm{cat}}}} \times \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ....... + \frac{1}{{14,000}}} \right) \approx 6 \times {I_{{\rm{cat}}}} \times \ln \left( {14,000} \right) = 6 \times \ln \left( {14,000} \right) \times {I_{{\rm{cat}}}}\)

But what we hear is best measured not by the intensity level, but by the decibel level. How do we go from one the other?

Most people are familiar with the decibel scale. The decibel level of a sound with an intensity of

*I*watts per square meter is
\({\rm{dB = 10}} \times {\rm{ }}\log \left( {\frac{I}{{{I_0}}}} \right)\) .

Using this conversion, we get

\[\begin{array}{l}

{\rm{dB = 10}} \times {\rm{log}}\left( {\frac{{6 \times \ln \left( {14,000} \right) \times {I_{{\rm{cat}}}}}}{{{I_0}}}} \right) = 10 \times \left( {\log \left( {6 \times \ln \left( {14,000} \right)} \right) + \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)} \right) = \\

10 \times 1.8 + 10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right) = 18 + 10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)

\end{array}\].

{\rm{dB = 10}} \times {\rm{log}}\left( {\frac{{6 \times \ln \left( {14,000} \right) \times {I_{{\rm{cat}}}}}}{{{I_0}}}} \right) = 10 \times \left( {\log \left( {6 \times \ln \left( {14,000} \right)} \right) + \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)} \right) = \\

10 \times 1.8 + 10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right) = 18 + 10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)

\end{array}\].

Recognize that \[10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)\] is just the decibel level of one cat from 1 meter away which the video says is 45 dB. All the cats in the above arrangement will give a decibel reading of (18+45) dB=63 dB.

Remember that an increase of 3 dB is a doubling of loudness, 18 dB is 6 doublings, so all the cats in the world meowing at once will sound like \({2^6} = 64\) cats 1 meter away.

I will leave it as an exercise for the reader to show that if you reduce the distance between cats to 0.5 meters, the dB level will 69 dB and will sound like 256 cats 1 meter away.

¹ I realize that \(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{{14000}} \approx \ln \left( {14000} \right) + \frac{1}{2}\) is a better approximation to the sum of the harmonic series (look at the geometry of the area under 1/x and the areas from the harmonic series), but that extra term of 0.5 doesn't change the final result appreciably.