# How to calculate sound intensity when given decibels

by DMOC
Tags: decibels, intensity, sound
 P: 100 1. The problem statement, all variables and given/known data When a person wears a hearing aid, the sound intensity level increases by 30.0 dB. By what factor does the sound intensity increase? 2. Relevant equations B = (10 dB) log (Ia/Ib) Ia = sound intensity Ib = threshold of human hearing (1.0 * 10^-12 W/m^2) 3. The attempt at a solution I used the equation above, substituting into it: 30 dB = (10 dB) log (Ia/1.0 * 10^-12 W/m^2) I divided both sides by 10 dB to get rid of the 10 dB on the right side of the equation. 3 = log (Ia/1.0 * 10^-12 W/m^2) Then I converted this into regular exponential form without logs: 10^3 = (I/1.0 * 10^-12 W/m^2) I get 1.0 * 10^-9 for I, but my answer key says the answer is 1000. This is 10^3, which I seem to have, but I still have an unknown - I - in the equation.
 PF Patron HW Helper P: 3,394 Your answer of I = 1.0 * 10^-9 is correct. But the question doesn't ask for I, it asks by what factor has I increased - presumably from 0 db which is the threshold of hearing. So divide your answer for I by the value of Ib.
HW Helper
P: 6,185
 Quote by DMOC 10^3 = (I/1.0 * 10^-12 W/m^2)
Right there is your answer. The wanted the factor, I/I0, not the intensity to give 30dB

P: 100

## How to calculate sound intensity when given decibels

So I got the right answer all along...it was just that I had to realize that the question asked for what (Ia/Ib) was equal to, not just Ia.

Thanks!

 Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 4 Introductory Physics Homework 1 Introductory Physics Homework 5 General Physics 7