Question

At a rock concert, a dB meter registered 127 dBwhen placed 2.2 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.

What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?

Express your answer to two significant figures and include the appropriate units.

How far away would the sound level be 87 dB ?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer 2

1). Power Output of the Speaker

Given:

• Sound level $L=127\text{dB}$

• Distance $r=2.2\text{m}$

• Reference intensity $I_0=1.0\times 10^{-12}{\text{W/m}}^2$

Steps:

1. Convert dB to Intensity $I$:

   $$L=10\log \left(\frac{I}{I_0}\right)$$$$127=10\log \left(\frac{I}{1.0\times 10^{-12}}\right)$$$$\frac{I}{1.0\times 10^{-12}}=10^{12.7}$$$$I=10^{12.7}\times 1.0\times 10^{-12}=10^{0.7}\approx 5.01{\text{W/m}}^2$$

2. Calculate Power $P$:

   $$I=\frac{P}{4\pi r^2}$$$$P=I\times 4\pi r^2=5.01\times 4\pi (2.2{)}^2$$$$P\approx 5.01\times 4\times 3.1416\times 4.84\approx 304\text{W}$$

Answer:
The power output of the speaker is 300 W  

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2). Distance for 87 dB Sound Level :-

Given:

• New sound level $L^{\mathrm{\prime }}=87\text{dB}$

• Power $P=304\text{W}$

Steps:

1. Convert dB to Intensity $I^{\mathrm{\prime }}$:

   $$87=10\log \left(\frac{I^{\mathrm{\prime }}}{1.0\times 10^{-12}}\right)$$$$\frac{I^{\mathrm{\prime }}}{1.0\times 10^{-12}}=10^{8.7}$$$$I^{\mathrm{\prime }}=10^{8.7}\times 1.0\times 10^{-12}=10^{-3.3}\approx 5.01\times 10^{-4}{\text{W/m}}^2$$

2. Calculate Distance $r^{\mathrm{\prime }}$:

   $$I^{\mathrm{\prime }}=\frac{P}{4\pi r^{\mathrm{\prime }2}}$$$$r^{\mathrm{\prime }2}=\frac{P}{4\pi I^{\mathrm{\prime }}}=\frac{304}{4\pi \times 5.01\times 10^{-4}}$$$$r^{\mathrm{\prime }2}\approx \frac{304}{6.28\times 10^{-3}}\approx 48,407{\text{m}}^2$$$$r^{\mathrm{\prime }}\approx \sqrt{48,407}\approx 220\text{m}$$

Answer:
The sound level would be 87 dB at a distance of 220 m  .

  

Answer 3

Absolutely, let's solve this problem step-by-step to get the correct answers.

1. Calculate the Sound Intensity at 2.2 meters

The sound level (β) in decibels (dB) is related to the sound intensity (I) by the following formula:

β = 10 * log₁₀(I / I₀)

where:

• β is the sound level in dB (127 dB)

• I is the sound intensity we want to find

• I₀ is the reference intensity (1.0 x 10⁻¹² W/m²)

Rearrange the formula to solve for I:

I = I₀ * 10^(β / 10)

Plug in the values:

I = (1.0 x 10⁻¹² W/m²) * 10^(127 / 10)

I = (1.0 x 10⁻¹² W/m²) * 10^12.7

I ≈ (1.0 x 10⁻¹² W/m²) * 5.01 x 10¹²

I ≈ 5.01 W/m²

2. Calculate the Power Output of the Speaker

Assuming uniform spherical spreading of sound, the intensity is related to the power (P) by:

I = P / (4πr²)

where:

• I is the intensity (5.01 W/m²)

• P is the power we want to find

• r is the distance from the source (2.2 m)

Rearrange the formula to solve for P:

P = I * 4πr²

Plug in the values:

P = 5.01 W/m² * 4π * (2.2 m)²

P ≈ 5.01 W/m² * 4π * 4.84 m²

P ≈ 304.7 W

Round to two significant figures:

P ≈ 300 W

3. Calculate the Distance for 87 dB Sound Level

First, find the intensity (I₂) at 87 dB:

I₂ = I₀ * 10^(β₂ / 10)

I₂ = (1.0 x 10⁻¹² W/m²) * 10^(87 / 10)

I₂ ≈ (1.0 x 10⁻¹² W/m²) * 5.01 x 10⁸

I₂ ≈ 5.01 x 10⁻⁴ W/m²

Now, use the power and intensity relationship:

I₂ = P / (4πr₂²)

Rearrange to solve for r₂:

r₂ = √(P / (4πI₂))

Plug in the values:

r₂ = √(304.7 W / (4π * 5.01 x 10⁻⁴ W/m²))

r₂ ≈ √(48347.5)

r₂ ≈ 219.88 m

Round to two significant figures:

r₂ ≈ 220 m

Answers:

1. The power output of the speaker is 300 W.

2. The distance at which the sound level would be 87 dB is 220 m.