Question

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) \(3.2 \mathrm{~dB}\) b) \(11 \mathrm{~dB}\) c) \(32 \mathrm{~dB}\) d) \(55 \mathrm{~dB}\) e) \(66 \mathrm{~dB}\)

Short answer

a) 30 dB b) 40 dB c) 50 dB d) 60 dB e) 66 dB

Step-by-step solution

Calculate the initial intensity of the sound

Use the formula \(L = 10 \log_{10}(\frac{I}{I_0})\), and solve it for intensity \(I\). We're given the initial sound level as \(80\mathrm{~dB}\) and the reference intensity \(I_0\) as \(10^{-12} \mathrm{~W/m^2}\). \(I = I_0 \times 10^\frac{L}{10} = 10^{-12} \times 10^\frac{80}{10} = 10^{-12} \times 10^8 = 10^{-4} \mathrm{~W/m^2}\)

Calculate the initial power of the speaker

Using the formula \(I = \frac{P}{4 \pi r^2}\), solve for \(P\). We're given the distance from the speaker \(r\) as \(3 \mathrm{~m}\). \(P = I \times 4 \pi r^2 = 10^{-4} \times 4 \pi(3)^2 = 10^{-4} \times 36\pi = 3.6\pi \times 10^{-3} \mathrm{~W}\)

Calculate the new (reduced) power of the speaker

We're given that the power of the speaker is reduced by a factor of \(25\). Multiply the initial power by \(\frac{1}{25}\) to get the new power: \(P_{new} = P \times \frac{1}{25} = 3.6\pi \times 10^{-3} \times \frac{1}{25} = \frac{3.6\pi}{25} \times 10^{-3} \mathrm{~W}\)

Calculate the new intensity of the sound

Use the formula \(I = \frac{P}{4 \pi r^2}\) with the new power. The distance from the speaker remains the same, \(3 \mathrm{~m}\). \(I_{new} = \frac{P_{new}}{4 \pi r^2} = \frac{\frac{3.6\pi}{25} \times 10^{-3}}{4 \pi (3)^2} = \frac{3.6\pi}{25} \times \frac{1}{36\pi} \times 10^{-4} = \frac{1}{25} \times 10^{-4} \mathrm{~W/m^2}\)

Calculate the new sound level

Use the formula \(L = 10 \log_{10}(\frac{I}{I_0})\) with the new intensity. The reference intensity \(I_0\) is still \(10^{-12} \mathrm{~W/m^2}\). \(L_{new} = 10 \log_{10}(\frac{I_{new}}{I_0}) = 10 \log_{10}(\frac{\frac{1}{25} \times 10^{-4}}{10^{-12}}) = 10 \log_{10}(25 \times 10^8) = 10 \log_{10}(10^2 \times 10^8) = 10 (2 + 8) = 100\) Since the new sound level is not an option in the given choices, we should double-check our calculations. Notice that we calculated an increase in sound level, which should not have been the case since the power was reduced. The error lies in calculating the new intensity. In Step 4, we actually calculated the ratio \(I_{new}/I_0\) instead of just \(I_{new}\). This means that we can skip directly to Step 5. Fixing this issue, our solution becomes:

(corrected): Calculate the new intensity of the sound

The sound intensity follows an inverse-square relationship with power. Since the power was reduced by a factor of 25, the new intensity should be: \(I_{new} = I / 25 = 10^{-4} / 25 = 4 \times 10^{-6} \mathrm{~W/m^2}\)

(corrected): Calculate the new sound level

Use the formula \(L = 10 \log_{10}(\frac{I}{I_0})\) with the corrected new intensity. The reference intensity \(I_0\) is still \(10^{-12} \mathrm{~W/m^2}\). \(L_{new} = 10 \log_{10}(\frac{I_{new}}{I_0}) = 10 \log_{10}(\frac{4 \times 10^{-6}}{10^{-12}}) = 10 \log_{10}(4 \times 10^6) = 10(6 + \log_{10}(4)) \approx 10(6 + 0.6) = 66\) So the new sound meter reading will be approximately \(\bold{66 \mathrm{~dB}}\), which is option e).