The noise in the school cafeteria is recorded at 50dB at 10:00. At 12:00 the noise is found to be 100dB. By what factor does the intensity of the sound increase at lunchtime?
The bel is the logarithm to the base 10 of one measurement of a physical quantity relative a reference level. Decibel is one tenth of a bel.
An increase of 1 decibel in the measurement of a physical quantity implies the quantity increasing by 10 times.
Let the reference for measuring the intensity of noise be N.
At 10:00, let the noise intensity in the cafeteria be N_10. This in terms of decibel is 50 dB
=> `10*log_10(N_10/N) = 50`
=> `(N_10/N) = 10^5`
=> `N_10 = 10^5*N`
At 12:00, the noise in terms of decibel is 100. Let the intensity of the sound be N_12
=> `10*log_10(N_12/N) = 100`
=> `(N_12/N) = 10^10`
=> `N_12 = 10^10*N`
`N_12/N_10 = 10^10/10^5`
The intensity of sound increased by a factor of `10^5` .
Let `I_1` be the intensity at 10:00,`I_2` the intensity at 12:00.
` `Note: The level of sound, measured in decibels (dB), is found by comparing the intensity of a sound (measured in watts per square meter) to the threshold of human hearing `I_0=10^(-12)W/m^2` .
Thus `L=10log_(10)(I/I_0)` .
(3) So the sound intensity increases by a factor of `I_2/I_1=10^22/10^17=10^5`