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.
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