12. The noise in the school cafeteria is recorded at 50 dB at 10:00 am. At noon, the noise is

found to be 100 dB. By what factor does the intensity of the sound increase at lunchtime?

1 answer

To determine the factor by which the intensity increases, we need to compare the intensities at 10:00 am and noon.

The intensity of sound is measured on a logarithmic scale, so we can use the formula:

dB = 10 * log10(I/I₀)

where dB is the decibel level, I is the intensity of the sound, and I₀ is a reference intensity.

Let's use the given decibel levels to calculate the two intensities:

50 dB = 10 * log10(I₁/I₀)

100 dB = 10 * log10(I₂/I₀)

To find the factor by which the intensity increases, we can divide these two equations:

(100 dB) / (50 dB) = (10 * log10(I₂/I₀)) / (10 * log10(I₁/I₀))

Simplifying:

2 = log10(I₂/I₀) / log10(I₁/I₀)

Since log10(I₂/I₀) / log10(I₁/I₀) represents the logarithm base 10 of a number, we can rewrite it as:

10^(log10(I₂/I₀) / log10(I₁/I₀))

The logarithmic property log(a/b) = log(a) - log(b) can be applied, giving:

10^(log10(I₂) - log10(I₀)) / (log10(I₁) - log10(I₀))

Since log10(I₀) is the same for both intensities, it can be canceled out:

10^(log10(I₂) - log10(I₀)) / (log10(I₁) - log10(I₀)) = 10^(log10(I₂) - log10(I₁))

Using the property a^(b - c) = a^b / a^c:

10^(log10(I₂) - log10(I₁)) = I₂ / I₁

Therefore, the factor by which the intensity increases is I₂ / I₁.

Substituting the given values:

I₂ / I₁ = 100 dB / 50 dB = 10.

Therefore, the intensity of the sound increases by a factor of 10 at lunchtime.