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?
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.