A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m from the speaker.Part A -What is the sound intensity at the position of the microphone?Express your answer in mW/m2and using two significant figures.I = _mW/m2Part B -What is the sound intensity level at the position of the microphone?Express your answer in dB using two significant figures.β= _dB

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

[tex]I = \frac{P}{A}[/tex]

[tex]I = \frac{P}{4\pi r^{2}}[/tex]

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

[tex]I = \frac{34}{4\pi (44)^{2}}[/tex]

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β [tex] = 10log_{10}\frac{I}{I_{0}}[/tex] dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β [tex] = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}[/tex]

β [tex] = 10log_{10} (1.4 * 10^{9})[/tex]

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

boffeemadrid boffeemadrid · Answer 2 · 2022-02-09T07:49:53+01:00

The sound intensity and sound intensity level is required.

The sound intensity is [tex]1.4\ \text{mW/m}^2[/tex]

The sound intensity level is [tex]91\ \text{db}[/tex]

Sound

P = Power = 34 W

r = Distance = 44 m

[tex]I_0[/tex] = Threshold of hearing = [tex]10^{-12}\ \text{W/m}^2[/tex]

A = Area = [tex]4\pi r^2[/tex]

Pressure divided by area gives sound intensity

Sound intensity is given by

[tex]I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{34}{4\pi \times44^2}\\\Rightarrow I=0.001397\ \text{W/m}^2=1.4\ \text{mW/m}^2[/tex]

Sound intensity level is given in the units of decibel (db)

Sound intensity level is given by

[tex]\beta=10\log\dfrac{I}{I_0}\\\Rightarrow \beta=10\log\dfrac{0.001397}{10^{-12}}\\\Rightarrow \beta=91.45\ \text{db}[/tex]

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