The problem you must solve is How many times louder is a 60

The problem you must solve is:

How many times louder is a 60 dB sound than a 20 dB sound?

How many times more energy is released by a magnitude 6 earthquake than by a magnitude 2 earthquake?

When you are 2 meters away from the speakers at a concert, the sound level is 130 db. How far away should you sit to reduce the level to 110 dB?

Answer all parts of the problem.

Show your justification for every step in your solution. Use clear, mathematically accurate language.

Label all numbers with the units they represent (e.g., 0.3048 ft/meter).

Clearly state your conclusions using complete English sentences (for example, “Jill needs to add 43 gallons of water to her pool”).

Solution

Answer:

(1) A 10dB increase in intensity means loudness is doubled.
    So we have
   20dB is 2 times louder than 10dB
30dB is 2 times louder than 20dB
40dB is 2 times louder than 30dB
50dB is 2 times louder than 40dB
60dB is 2 times louder than 50dB

So we have 60dB is 2x2x2x2 times louder than 20dB.
2x2x2x2x = 2^4 = 16
So 60dB is 16 times louder than20dB.

(2) Given the magnitude (M), we compute the released energy (E) using the following formula:

     E = (2.5 x 10^4) x 10^(1.5xM)

     Energy is measured in joules.

     Magnitudes have no units.

If M=6 then E=(2.5 x 10^4) x 10^(1.5x6)

                      =(2.5 x 10^4) x 10^(9)

                      =(2.5 x 10^13) joules

If M=2 then E=(2.5 x 10^4) x 10^(1.5x2)

                      =(2.5 x 10^4) x 10^(3)

                     =(2.5 x 10^7) joules

Now consider (2.5 x 10^13)/(2.5 x 10^7)=10^6

therefore 10^6 times more energy is released by a magnitude 6 earthquake than by a magnitude 2 earthquake.

(3) Using the fact that the intensity decreases as inverse r square. r is the distance.
that is I = I0/r^2.
at r= 2meters , I= 130,
This gives

I0= 520 dB
Now at I=110 dB. Therefore
110 = 520/r^2,
r^2= 520/110=52/11=4.72.
r= 2.174m