As people sing in church, the sound level everywhere inside is 122 dB. No sound

Question

As people sing in church, the sound level everywhere inside is 122 dB. No sound is transmitted through the massive walls, but all the windows and doors are open on a summer morning. Their total area is 22.5 m2. (Enter only the answers in the input boxes separately given.)

(a) How much sound energy is radiated in 26.5 min?

1. ( Enter in box 1) J

(b) Suppose the ground is a good reflector and sound radiates uniformly in all horizontal and upward directions. Find the sound level 1.4km away.

2. ( Enter in box 2)

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.) (Enter only the answers in the input boxes separately given.)

(c) Find the wavelength of the initial note.

3. ( Enter in box 3) m

(d) Find the wavelength of the final note.

4. (Enter in box 4) m

(e) Assume the choir sings the melody with a uniform sound level of 87.0 dB. Find the pressure amplitude of the initial note.

5. ( Eneter in box 5) Pa

(f) Find the pressure amplitude of the final note.

6. ( Enter in box 6) Pa

(g) Find the displacement amplitude of the initial note.

7. ( Enter in box 7) m

Just do G

Explanation / Answer

intensity in dB(ID)=10log(I/I0)

Since Intensity is proportional to (amplitude)2

ID=20*log(P/P0) ; where P0-threshold frequency = 2*10-5 Pa

Since the intensity is constant which is 87dB

P= 0.448 Pa

Now,, P= (vdw)*s; where P-pressure amplitude, v-velocity. s-distance amplitude, w-angular frequency. d-density

and w=2*pi*f

So, the answer s= 1.18*10-6 m

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