The combined sound pressure level measured at a point in a production bench due to one dumper and one shovel is 95 dB(A). If the sound pressure level of shovel alone is 90 dB(A), t... The combined sound pressure level measured at a point in a production bench due to one dumper and one shovel is 95 dB(A). If the sound pressure level of shovel alone is 90 dB(A), the sound pressure level of the dumper alone, in dB(A), at the same point is _______ (round off up to 2 decimals)
Understand the Problem
The question is asking us to determine the sound pressure level of a dumper given the combined sound pressure level of a dumper and a shovel and the sound pressure level of the shovel alone. We will use the properties of sound levels in decibels (dB).
Answer
The sound pressure level of the dumper alone is approximately $93.37 \, \text{dB(A)}$.
Answer for screen readers
The sound pressure level of the dumper alone is approximately $93.37 , \text{dB(A)}$.
Steps to Solve
- Understanding the Sound Level Equation
The sound level in decibels (dB) is not additive in the linear sense. Instead, we convert levels back to the intensity and then combine them. The formula for sound pressure levels is given by:
$$ L = 10 \log_{10} \left( \frac{I}{I_0} \right) $$
where $I$ is the intensity and $I_0$ is the reference intensity.
- Convert dB Levels to Intensities
First, we will convert the given sound levels of the shovel and combined levels to intensities. The sound pressure level of the shovel is 90 dB(A) and the combined level is 95 dB(A).
Using the formula to find intensity for the shovel:
$$ I_s = I_0 \times 10^{(L_s / 10)} $$
where $L_s = 90 , \text{dB(A)}$.
Thus,
$$ I_s = I_0 \times 10^{(90 / 10)} = I_0 \times 10^9 $$
Now for the combined level:
$$ I_c = I_0 \times 10^{(L_c / 10)} $$
where $L_c = 95 , \text{dB(A)}$.
Thus,
$$ I_c = I_0 \times 10^{(95 / 10)} = I_0 \times 10^{9.5} $$
- Use the Relationship Between Intensities
The total intensity of the system when both the dumper and the shovel are present is given by:
$$ I_c = I_d + I_s $$
where $I_d$ is the intensity of the dumper. Therefore, rewriting this gives:
$$ I_d = I_c - I_s $$
Substituting the expressions we found earlier:
$$ I_d = I_0 \times 10^{9.5} - I_0 \times 10^9 $$
Factoring out $I_0$ gives:
$$ I_d = I_0 \left( 10^{9.5} - 10^9 \right) $$
- Calculate the dB Level of the Dumper Alone
Using the equation for sound level, we can find the dB level of the dumper:
$$ L_d = 10 \log_{10} \left( \frac{I_d}{I_0} \right) $$
Substituting in $I_d$:
$$ L_d = 10 \log_{10} \left( 10^{9.5} - 10^9 \right) $$
- Simplify and Calculate the dB Level
Calculating the values inside the log we have:
$$ L_d = 10 \log_{10} \left( 10^{9.5} (1 - 10^{-0.5}) \right) $$
So, simplifying it gives:
$$ L_d = 10 \left( 9.5 + \log_{10} (1 - 10^{-0.5}) \right) $$
Now calculate $10^{-0.5} \approx 0.3162$, so:
$$ L_d \approx 10 \left( 9.5 + \log_{10} (0.6838) \right) $$
Calculate $\log_{10} (0.6838) \approx -0.163$:
$$ L_d \approx 10 \left( 9.5 - 0.163 \right) $$
Thus,
$$ L_d \approx 10 \times 9.337 \approx 93.37 , \text{dB(A)} $$
The sound pressure level of the dumper alone is approximately $93.37 , \text{dB(A)}$.
More Information
This calculation illustrates how sound pressure levels cannot be simply combined arithmetically but need to be converted to intensities first. The resulting level shows the impact of both machines at the measured point.
Tips
- Ignoring the conversion from dB to intensity: Always remember that sound levels in dB are logarithmic, so direct addition is incorrect.
- Failing to correctly factor out $I_0$ when calculating combined intensities.
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