Question

The loudness of the sound in a concert hall and in a football field are 80 dB and 85 dB respectively. Find the ratio of the sound intensities in the football field to the concert hall.

Answer 1

85db/80db = 16/17
so every 16db in concert hall will have 17db in football field

Answer 2

The sound intensity on the football field is greater than the sound intensity in the concert hall by (85 - 80)dB.

Answer 3

Sound Intensity is proportional to the power in watts
Relation between decibels and power
\[ db= 10 \log_{10} P\]
P= power in watts
here let power P1 be in concert hall, P2 be the power in the football field
Find P2/P1

Answer 4

@kryton1212 Do you get this?

Answer 5

yes =] thank you

Answer 6

@kryton1212 Can you do the next part?

Answer 7

wait a moment

Answer 8

Sure

Answer 9

sorry I cannot get it...

Answer 10

The sound intensity on the football field is 5dB greater than the sound intensity in the concert hall. The measurements of the two intensities are relative to the same reference level and are logarithmic. Therefore subtraction is required in this case.

Answer 11

I think the question is asking the ratio not the difference.

Answer 12

The result of 5dB is a ratio.

Answer 13

but the answer is 3.2:1

Answer 14

Yeah, 5 db is ratio. You could convert that to power, that will also be the ratio.
\[5=10\log_{10} P\]
P is the ratio

Answer 15

(85-80)dB=10 log P which P is the ratio of the sound intensities

Answer 16

yes

Answer 17

A power ratio of 5dB is equivalent to a power ratio of 3.2:1

Answer 18

why?

Answer 19

Subtraction of dbs, is equivalent to division of powers. That's why it's a ratio

Answer 20

5=10logP
and then?

Answer 21

\\[5=10\log_{10} P\]
\[\frac{5}{10}=\log_{10} P\]
Take anti log both sides
\[10^{\frac{5}{10}}=P\]
\[10^{\frac 1 2}=P\]

Answer 22

\[\log_{} P=\frac{5}{10}\]
\[10^{0.5}=3.16\]

Answer 23

thank you. I know now