2The toadfish makes use of resonance in a closed tube to pro

2.The toadfish makes use of resonance in a closed tube to produce very loud sounds. The tube is its swim bladder, used as an amplifier. The sound level of this creature has been measured as high as 90 dB.

(a) Calculate the intensity of the sound wave emitted.
(b) What is the intensity level if seven of these fish try to imitate seven frogs by saying \"Budweiser\" at the same time?

3.An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 441 Hz when the train is approaching him and 439 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find

4.An outside loudspeaker (considered a small source) emits sound waves with a power output of 80 W.

(a) Find the intensity 7.5 m from the source.
(b) Find the intensity level in decibels at that distance.

Solution

(2) Sound intensity level in dB of a sound at intensity I is defined as:
L = 10log( I/I )
with I = 10¹²W/m² threshold of hearing

Therefore
I = I 10^( L/10 )

So a sound with a level of 75dB has an intensity of
I = 10¹²W/m² 10^( 90/10 )


Because sound intensity level is a logarithmic scale, multiplying the sound intensity by some factor corresponds to an addition of some other factor to the level.
Let L\' be the SIL associated with r times of original intensity I and level L, i.e.:
L\' = 10log( r I/I)
because log(ab) = log(a) + log(b)
L\' = 10log(r) + 10log( I/I )
<=>
L\' = 10log(r) + L

So seven fish sounding together cause a sound intensity level of:
L\' = 10log(7) + 90 (db)

(3) From the stationary listener Doppler eqn we have

f\' = f*v/(v + vs)

When the source is approaching vS is negative and assuming v (speed of sound) = 343

We have 441 = f*343/(343 - vS)

or 441*(343 - vS) = f*343

151263 - 441vS = 343f                           (eqn 1)

For the case where the train has past vS is positive

We have 439 = f*343/(343 + vS)

or 439*(343 + vS) = f*343

150577+ 439vS = 343f                               (Eqn 2)
From eqn1 & eqn 2 we get

151263 - 441vS= 150577+ 439vS

686=880vs

vs=0.77 m/sec

(4) Dividing the power by the surface area of a sphere at a distance R = 10.5 meters.

(a)80 (4R²) = 80 [4(7.5)²] = 0.113 W/m²

(b) db(s) = 10log[(0.113) (1 ×10¹²)] = db(s)