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Chapter 16: Problem 6

A sound meter placed \(3 \mathrm{~m}\) from a speaker registers a sound level of \(80 \mathrm{~dB}\). If the volume on the speaker is then turned down so that the power is reduced by a factor of 25 , what will the sound meter read? a) \(3.2 \mathrm{~dB}\) c) \(32 \mathrm{~dB}\) e) \(66 \mathrm{~dB}\) b) \(11 \mathrm{~dB}\) d) \(55 \mathrm{~dB}\)

Short Answer

Expert verified
Answer: The new sound level will be 66 dB.

Step by step solution

01

Analyze the given information

We are given: - Initial sound level, \(L_1 = 80 \mathrm{~dB}\) - Initial distance from the speaker, \(r_1 = 3 \mathrm{~m}\) - Power is reduced by a factor of 25. We need to find the new sound level when power is reduced.
02

Recall the formula for sound level in dB

The formula for sound level in dB is as follows: $$L = 10 \log_{10}\left(\frac{I}{I_0}\right)$$ Where: - \(L\) is the sound level in dB - \(I\) is the intensity of the sound at the given point - \(I_0\) is the reference intensity (\(10^{-12}\) W/m²)
03

Calculate the initial sound intensity

First, we'll find the initial sound intensity by using the given sound level. $$80 = 10 \log_{10}\left(\frac{I_1}{I_0}\right)$$ Now, solving for \(I_1\): $$I_1 = I_0 \cdot 10^{(80/10)}$$
04

Calculate the new sound intensity

The power is reduced by a factor of 25. Therefore, the new intensity, \(I_2\), will also be reduced by a factor of 25. $$I_2 = \frac{I_1}{25}$$
05

Calculate the new sound level

Now, we'll use the formula for sound level in dB to find the new sound level, \(L_2\). $$L_2 = 10 \log_{10}\left(\frac{I_2}{I_0}\right)$$ Plug in the value of \(I_2\) from Step 4: $$L_2 = 10 \log_{10}\left(\frac{\frac{I_1}{25}}{I_0}\right)$$ Now, plug in the value of \(I_1\) found in Step 3: $$L_2 = 10 \log_{10}\left(\frac{\frac{I_0 \cdot 10^{(80/10)}}{25}}{I_0}\right)$$
06

Simplify and solve for the new sound level

Simplify the expression by canceling the term \(I_0\) and the factor of 25: $$L_2 = 10 \log_{10}\left(10^{(80/10 - \log_{10}{25})}\right)$$ Now solve for \(L_2\): $$L_2 = 80 - 10(\log_{10}{25}) \approx 66$$
07

Find the correct answer

By comparing the obtained value of \(L_2\) to the given answer choices, we find that the correct answer is: e) \(66 \mathrm{~dB}\)

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Most popular questions from this chapter

A bugle can be represented by a cylindrical pipe of length \(L=1.35 \mathrm{~m} .\) Since the ends are open, the standing waves produced in the bugle have antinodes at the open ends, where the air molecules move back and forth the most. Calculate the longest three wavelengths of standing waves inside the bugle. Also calculate the three lowest frequencies and the three longest wavelengths of the sound that is produced in the air around the bugle.

A standing wave in a pipe with both ends open has a frequency of \(440 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(660 \mathrm{~Hz}\) a) Determine the fundamental frequency. b) How long is the pipe?

The sound level in decibels is typically expressed as \(\beta=10 \log \left(I / I_{0}\right),\) but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on the amplitude squared, so the expression is \(\beta=20 \log \left(P / P_{0}\right),\) where \(P_{0}\) is the smallest pressure difference noticeable by the ear: \(P_{0}=2.00 \cdot 10^{-5} \mathrm{~Pa}\). A loud rock concert has a sound level of \(110 . \mathrm{dB}\), find the amplitude of the pressure wave generated by this concert.

Compare the intensity of sound at the pain level, \(120 \mathrm{~dB}\), with that at the whisper level, \(20 \mathrm{~dB}\).

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