Question

As the people sing in church, the sound level everywhere inside is 101 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 22.0 m² . (b) Suppose the ground is a good reflector and sound radiates from the church uniformly in all horizontal and upward directions. Find the sound level 1.00 km away.

Answer 1

The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

What do you mean by sound radiates?

Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.

(Lw) = 10·log (W/Wo) dB

Given:

sound level,  $\beta= 101 dB$

Area, A = $22\;m^{2}$

Time, $\triangle t = 20\;min=1200\;s$

Intensity, $I=1\times 10^{-12}\;W/m^{2}$

$r=1\;km=1000\;m$

(a)

We know that, Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Solving the above equation for sound intensity,

$I=I_{o} \times 10^{\frac{\beta}{10} }$

$I=1 \times 10^{-12} \times 10^{\frac{101}{10} }$

$I=0.0126\;W/m^{2}$

Therefore, The sound energy is,

$E=P\times \triangle t$

Substitute $P=I \times A$ in the above equation,

$E=I \times A \times \triangle t$

$E=0.0126 \times 22 \times 1200$

$E=332.6\;J$

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

$A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi$

Substitute the known value in the above equation ,

$A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi$

$A_{hemisphere} = 6283185\;m^{2}$

Sound Intensity is,

$I = \frac{P}{A_{hemisphere}}$

Substitute $P=I \times A$ in the above equation,

$I = \frac{I \times A}{A_{hemisphere}}$

Substitute the known value in the above equation,

$I = \frac{0.0126 \times 22}{6283185}$

$I = 4.4 \times 10^{-8}\;W/m^{2}$

Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Substitute the known value in the above equation,

$\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )$

$\beta=46.4\;dB$

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