Question

(a) The power of sound from the speaker of a radio is 20 mW. By turning the knob of volume control the power of sound is increased to 400 mW, What is the power increase in dB as compared to original power? (b) How much more intense is an 80 dB sound than a 20 dB whisper?

Answer 1

a. As intensity is power per unit area, for a given source `PmuI`. If `L_1` and `L_2` are the initial and final loudness levels, Then we have
`L_2-L_1=10log((I_2)/(I_1))`
The increase in loudness level is
`triangleL=10log((P_2)/(P_1))=10log((400)/(20))`
`impliestriangleL=10[log20]=13dB`
b. Increase in loudness level of sound is
`L_2-L_2=10log((I_2)/(I_1))`
`80-20=10log((I_2)/(I_1))`
`implies6=log((I_2)/(I_1))`
`implies((I_2)/(I_1))=10^6`