The intensity of a sound at a point $P$ is inversely proportional to the square of the distance from $P$ to the source $S$. When the distance $PS = x$ m, the intensity is 60 dB. When the distance is decreased by 20%, find
a) the new distance between P and S in terms of $x$,
b) the new intensity of the sound at P,
c) the percentage increase in the intensity of the sound at P.
I got the following and would like to check if my answer is right.
- a) $0.8x^2$
- b) $1.2\times60 = 72$ dB
- c) $(72-60/60)\times100 = 20\%$
No, this isn't correct.
a) if the distance was $x\,\mathrm{m}$ and has been decreased by $20\%$, the new distance is just $0.8x\,\mathrm{m}$.
b) The new square of the distance is $(0.8x)^2\,\mathrm m^2$. We know that the intensity is inversely proportional to the square of the distance. When this was $x^2$, the intensity was $60$ dB, so now it is $60/(0.8)^2$ dB.
Once you have the right answer for b), you will get the right answer for c).