Question

a noisy workplace has a noise level of 90 db. you want to reduce this to a more comfortable 75 db. by what factor must the power of the noise source be reduced? (that is, what is the ratio of the new power to the old?)

Answer 1

To reduce the noise level from 90 dB to 75 dB, the power of the noise source must be reduced by a factor of approximately 31.62, meaning the new power should be 1/31.62 of the old power.

Step-by-step explanation:

To determine by what factor the power of a noise source must be reduced to decrease the sound intensity level from 90 dB to 75 dB, we can use the decibel formula:

Level (dB) = 10 log10(I/I0)

where I is the sound intensity and I0 is the reference sound intensity. This formula relates the intensity of the sound to its level in decibels (dB). We know that every 10 dB change corresponds to a factor of 10 change in sound intensity. Thus, a change from 90 dB to 75 dB is a 15 dB decrease.

To find the factor by which the power must be reduced, we divide the decibel difference by 10:

Factor = 10(15/10)

= 101.5

This result means the power of the noise source must be reduced by a factor of approximately 31.62 (since 101.5 is roughly 31.62).

The ratio of the new power to the old power is therefore:

New Power / Old Power = 1 / 31.62