Question

1. Suppose that a single guitar player makes noise at 90 dB.

(a) If two guitar players play at the same time, what is the noise level in dB? What is the sound intensity in W/m2 ?

(b) How many guitar players would have to play in order for the sound intensity to be 110 dB?

Answer 1

For a single guitar player noise level = 90 dB

Intensity of one guitar will be

SL = 10*log (I/I0)

I0 = Human threshold intensity = 10^-12 W/m^2

I = intensity of sound produced by 1 guitar

SL = Intensity level of 1 guitar = 90 dB

So,

90 = 10*log (I/10^-12)

log (I/10^-12) = 90/10

I/10^-12 = 10^9

I = 10^9*10^-12 = 10^-3 W/m^2

Now intensity of 2 guitars will be

I1 = 2*I = 2*10^-3 W/m^2

Now Intensity level of 2 guitars will be

SL1 = 10*log (I1/I0)

SL1 = 10*log (2*10^-3/10^-12) = 10*log (2*10^9)

SL1 = 93.01 dB = noise level in dB for two guitar levels

Sound intensity = 2*10^-3 W/m^2

Part B.

Suppose n guitars produce sound intensity to be 110 dB, then intensity of n guitars (I_n) will be:

110 = 10*log (I_n/10^-12)

log (I_n/10^-12) = 110/10

I_n/10^-12 = 11

I_n = 10^11*10^-12 = 10^-1 W/m^2

Now intensity of 1 guitar = 10^-3 W/m^2

Intensity of n guitars = n*I = I_n

n = I_n/I

n = 10^-1/10^-3 = 100

100 guitar players would have to play to produce sound intensity of 110 dB