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A crying baby emits sound with an intensity of 8.0 × 10-8 w/m2. calculate a reasonable estimate for the intensity level from a set of quintuplets (five babies), all crying simultaneously at the same place? the lowest detectable intensity is 1.0 × 10-12 w/m2.

2 months ago

Solution 1

Guest #4094

2 months ago

Sound intensity of 1 baby, I = 8*10^-8 W/m^2

The sound heard should be higher by:

10*log (n) where for 5 babies, n = 5. Then
10*log (n) = 10*log (5) ≈ 7 dB

Also give is the reference sound, Io = 1.0*10^-12 W/m^2

Therefore,
Sound intensity, L1 = 10*log (I/I1) = 10*log [(8*10^-8)/(1*10^-12)] ≈ 49 dB
Therefore, total intensity for the five babies is:

Total intensity = 49+7 = 56 dB

Solution 2

Guest #4095

2 months ago

The intensity level from a set of quintuplets (five babies) : 56 dB

Further explanation

Wave intensity is the power of a wave that is moved through a plane of one unit that is perpendicular to the direction of the wave

Can be formulated

$\rm I=\dfrac{P}{A}$

I = intensity, W m⁻²

P = power, watt

A = area, m²

The farther the distance from the sound source, the smaller the intensity

$\rm \dfrac{I_2}{I_1}=\dfrac{(r_1)^2}{(r_2)^2}$

So the intensity is inversely proportional to the square of the distance from the source

$\rm I\approx \dfrac{1}{r^2}$

Intensity level (LI) can be formulated

$\rm LI=10\:log\dfrac{I}{I_o}$

Io = 10⁻¹²

For the level of intensity of several sound sources as many as n pieces can be formulated:

LIn = LI1 + 10 log n

The intensity level of 1 baby is

$\rm LI=10\:log\dfrac{8.10^{-8}}{10^{-12}}$

LI = 10 log 8.10⁴

LI = 49

The intensity level of 5 babies :

LI5 = LI + 10 log n

LI5 = 49 + 10 log 5

LI5 = 49 + 7

LI5 = 56

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