speaker and microphone urgent help

Question

. A speaker emits isotropic sound waves at a frequency of 3x103 Hz and a power of 25 Watts. A small microphone has an area of 0.72 cm2 and is located 4.12 meters from the point source. The speed of sound is 343 m/s.

a. What is the sound intensity at the microphone?

b. What is the power intercepted by the microphone?

c. What is the sound intensity level (in dB) at the microphone from this speaker?

d. What would be the sound intensity level (in dB) at the microphone if the speaker doubled its power?

e. If a second speaker is placed a distance 4.86 meters away from the microphone, what will happen to the sound intensity level at the microphone? Will it increase, decrease, or stay the same? Explain.

Answer 1

Given:

1. Speaker power: P = 25 W
2. Frequency: f = 3 × 10³ Hz (not actually needed for these parts)
3. Distance from speaker to microphone: r = 4.12 m
4. Microphone area: A = 0.72 cm²
5. Speed of sound: v = 343 m/s (not needed here)
6. Reference intensity: I₀ = 1.0 × 10⁻¹² W/m²
7. Temperature/air properties not needed

Note:

Convert area to m²:

A = 0.72 cm² = 0.72 × (10⁻² m)² = 0.72 × 10⁻⁴ m² = 7.2 × 10⁻⁵ m²

a) Sound intensity at the microphone

For an isotropic point source (spreading equally in all directions):

I = P / (4πr²)

I = 25 W / [4π(4.12 m)²]

I ≈ 25 / [4π × 16.9744]

I ≈ 25 / 213.4

I ≈ 0.117 W/m²

Answer (a):

I ≈ 0.117 W/m²

b) Power intercepted by the microphone

The power intercepted is:

P_mic = I × A

P_mic = (0.117 W/m²) × (7.2 × 10⁻⁵ m²)

P_mic ≈ 8.44 × 10⁻⁶ W

Answer (b):

P_mic ≈ 8.4 × 10⁻⁶ W

c) Sound intensity level (in dB) at the microphone

Sound intensity level (β) in decibels:

β = 10 log₁₀(I / I₀)

Take I₀ = 1.0 × 10⁻¹² W/m²:

β = 10 log₁₀(0.117 / 1.0 × 10⁻¹²)

β = 10 log₁₀(1.17 × 10¹¹)

β ≈ 10 × 11.07

β ≈ 111 dB

(Using a calculator gives about 110.7 dB.)

Answer (c):

β ≈ 1.11 × 10² dB (≈ 111 dB)

d) Sound intensity level if the speaker doubles its power

If the power doubles (from 25 W to 50 W):

1. Intensity at the same distance also doubles: I' = 2I (because intensity is proportional to power)
2. New level β' is:

β' = 10 log₁₀(I' / I₀)

= 10 log₁₀(2I / I₀)

= 10 [log₁₀(I / I₀) + log₁₀(2)]

= β + 10 log₁₀(2)

Since log₁₀(2) ≈ 0.301:

10 log₁₀(2) ≈ 3.0 dB

So:

β' ≈ 110.7 dB + 3.0 dB ≈ 113.7 dB

Answer (d):

The sound intensity level increases by about 3 dB to ≈ 114 dB

e) Second speaker at 4.86 m: what happens to the sound intensity level?

Now you have two speakers contributing sound intensity at the microphone.

Intensity from speaker 1 (at 4.12 m):

I₁ ≈ 0.117 W/m² (from part a)

Intensity from speaker 2 (same power, but at r₂ = 4.86 m):

I₂ = P / (4πr₂²)

= 25 / [4π(4.86)²]

≈ 25 / 297

≈ 0.084 W/m²

Total intensity (assuming incoherent sources so intensities add):

I_total = I₁ + I₂

≈ 0.117 + 0.084

≈ 0.201 W/m²

New level:

β_total = 10 log₁₀(I_total / I₀)

= 10 log₁₀(0.201 / 1.0 × 10⁻¹²)

≈ 10 log₁₀(2.01 × 10¹¹)

≈ 10 × 11.30

≈ 113 dB

So compared to the original ≈111 dB, the level increases by a couple of dB.

Answer (e):

The sound intensity level at the microphone increases. Adding a second speaker adds extra intensity (energy per area), so the total intensity is larger and thus the dB level is higher. It will not stay the same and definitely does not decrease.

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Symbol explanations:

1. P = power of the source (Watts, W)
2. r = distance from the sound source to the point of interest (meters, m)
3. I = sound intensity (W/m²), power per unit area
4. A = area of the microphone (m²)
5. P_mic = power intercepted by the microphone (W)
6. I₀ = reference intensity, usually 1.0 × 10⁻¹² W/m²
7. β = sound intensity level in decibels (dB)
8. π = pi ≈ 3.14159
9. log₁₀ = base-10 logarithm