Part A A concert loudspeaker suspended high off the ground emits 29.0 W of sound power. A small microphone with a 0.600 cm² area is 55.0 m from the speaker. What is the sound intensity at the position of the microphone? Express your answer with the appropriate units. 7.63×10-4 W m Submit Previous Answers v Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part B How much sound energy impinges on the microphone each second? Express your answer with the appropriate units. ? J -8 4.578 • 10 None
The Correct Answer and Explanation is:
Part A: Sound Intensity at the Microphone
We are given:
- Power of the loudspeaker: P=29.0P = 29.0P=29.0 W
- Distance from the loudspeaker: r=55.0r = 55.0r=55.0 m
- Area of the microphone: A=0.600A = 0.600A=0.600 cm² = 6.00×10−66.00 \times 10^{-6}6.00×10−6 m²
Step 1: Calculate the Sound Intensity
Sound intensity (III) is defined as power per unit area. Assuming the sound spreads uniformly in all directions, we use the formula: I=PAsphereI = \frac{P}{A_{\text{sphere}}}I=AsphereP
where AsphereA_{\text{sphere}}Asphere is the surface area of a sphere centered at the loudspeaker: Asphere=4πr2A_{\text{sphere}} = 4\pi r^2Asphere=4πr2
Substituting the given values: Asphere=4π(55.0)2A_{\text{sphere}} = 4\pi (55.0)^2Asphere=4π(55.0)2 Asphere=4π(3025)≈37994.57 m2A_{\text{sphere}} = 4\pi (3025) \approx 37994.57 \text{ m}^2Asphere=4π(3025)≈37994.57 m2
Now, calculating the intensity: I=29.037994.57≈7.63×10−4 W/m2I = \frac{29.0}{37994.57} \approx 7.63 \times 10^{-4} \text{ W/m}^2I=37994.5729.0≈7.63×10−4 W/m2
Thus, the sound intensity at the position of the microphone is 7.63×10−47.63 \times 10^{-4}7.63×10−4 W/m².
Part B: Sound Energy Impinging on the Microphone Each Second
The sound energy per second that reaches the microphone is given by: Pmicrophone=I×AP_{\text{microphone}} = I \times APmicrophone=I×A
where:
- I=7.632×10−4I = 7.632 \times 10^{-4}I=7.632×10−4 W/m² (using full unrounded value)
- A=6.00×10−6A = 6.00 \times 10^{-6}A=6.00×10−6 m²
Substituting: Pmicrophone=(7.632×10−4)×(6.00×10−6)P_{\text{microphone}} = (7.632 \times 10^{-4}) \times (6.00 \times 10^{-6})Pmicrophone=(7.632×10−4)×(6.00×10−6) Pmicrophone=4.58×10−9 WP_{\text{microphone}} = 4.58 \times 10^{-9} \text{ W}Pmicrophone=4.58×10−9 W
Since power (W) is energy per second (J/s), this means that 4.58×10−94.58 \times 10^{-9}4.58×10−9 J of energy impinges on the microphone each second.
Explanation
Sound intensity is a measure of how much power passes through a given area. Since the loudspeaker emits sound in all directions, the power is distributed over a growing spherical surface area as the sound travels outward. This means that at a larger distance, the intensity decreases because the same amount of power is spread over a much larger area.
To determine how much sound energy reaches the microphone, we used the concept that energy per second is equal to intensity multiplied by the surface area exposed. Since the microphone is very small compared to the spherical wavefront of the sound, only a tiny fraction of the total sound power reaches it.
These calculations are crucial in fields like acoustics, audio engineering, and physics, where understanding sound energy distribution helps in designing effective sound systems, microphones, and noise control measures.
