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Chapter 12: Problem 62

When playing fortissimo (very loudly), a trumpet emits sound energy at a rate of \(0.800 \mathrm{W}\) out of a bell (opening) of diameter \(12.7 \mathrm{cm} .\) (a) What is the sound intensity level right in front of the trumpet? (b) If the trumpet radiates sound waves uniformly in all directions, what is the sound intensity level at a distance of \(10.0 \mathrm{m} ?\)

Short Answer

Expert verified
Answer: The sound intensity level right in front of the trumpet is 118 dB, and the sound intensity level at a distance of 10 meters from the trumpet is 88 dB.

Step by step solution

01

Calculate the area of the trumpet opening

We are given the diameter of the trumpet opening, so we need to calculate its area using the formula for the area of a circle: \(A = \pi r^2\), where \(r\) is the radius, which is half of the diameter. Diameter = 12.7 cm Radius = \(\frac{12.7}{2} = 6.35\) cm Now, we can find the area: \(A = \pi (6.35)^2 \approx 126.7 \ \text{cm}^2\)
02

Calculate the sound intensity near the opening

The formula for sound intensity is \(I = \frac{P}{A}\), where \(I\) is the intensity, \(P\) is the power (0.800 W), and \(A\) is the area we've calculated. So, \(I = \frac{0.800 \, \text{W}}{126.7 \, \text{cm}^2} \approx \frac{0.800 \, \text{W}}{0.01267 \, \text{m}^2} = 63.1 \, \frac{\text{W}}{\text{m}^2}\)
03

Calculate the sound intensity level near the opening

Using the formula for sound intensity level, \( \beta = 10 \log_{10}(\frac{I}{I_0})\), where \(I_0 = 10^{-12} \, \frac{\text{W}}{\text{m}^2}\) is the reference intensity. \(\beta = 10 \log_{10}(\frac{63.1 \, \frac{\text{W}}{\text{m}^2}}{10^{-12} \, \frac{\text{W}}{\text{m}^2}}) \approx 118 \, \text{dB}\) So the sound intensity level right in front of the trumpet is 118 dB.
04

Calculate the sound intensity at a distance of 10 meters

We will use the inverse square law to calculate the intensity at a distance of 10 meters. The formula is \(I_r = \frac{P}{4 \pi r^2}\), where \(I_r\) is the intensity at a distance \(r\), \(P\) is the power, and \(r\) is the distance. \(I_r = \frac{0.800 \, \text{W}}{4 \pi (10.0 \, \text{m})^2} = \frac{0.800 \, \text{W}}{400 \pi \, \text{m}^2} \approx 6.36 \times 10^{-4} \, \frac{\text{W}}{\text{m}^2}\)
05

Calculate the sound intensity level at a distance of 10 meters

Using the same formula for sound intensity level as in step 3, we get: \(\beta_r = 10 \log_{10}(\frac{6.36 \times 10^{-4} \, \frac{\text{W}}{\text{m}^2}}{10^{-12} \, \frac{\text{W}}{\text{m}^2}}) \approx 88 \, \text{dB}\) So the sound intensity level at a distance of 10 meters from the trumpet is 88 dB.

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Most popular questions from this chapter

A bat emits chirping sounds of frequency \(82.0 \mathrm{kHz}\) while hunting for moths to eat. If the bat is flying toward the moth at a speed of $4.40 \mathrm{m} / \mathrm{s}\( and the moth is flying away from the bat at \)1.20 \mathrm{m} / \mathrm{s},$ what is the frequency of the sound wave reflected from the moth as observed by the bat? Assume it is a cool night with a temperature of \(10.0^{\circ} \mathrm{C} .\) [Hint: There are two Doppler shifts. Think of the moth as a receiver, which then becomes a source as it "retransmits" the reflected wave. \(]\)
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