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?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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`

= `10^5`

The intensity of sound increased by a factor of `10^5` .

embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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)` .

(1) `50=10log_(10)(I_1/I_0)`




(2) `100=10log_(10)(I_2/I_0)`




(3) So the sound intensity increases by a factor of `I_2/I_1=10^22/10^17=10^5`

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