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 _________. Read more

Steps to Solve

1. Understand the Levels in dB The sound pressure levels in decibels (dB) use a logarithmic scale. The combined sound pressure level ($L_{combined}$) of two sources can be calculated using the formula: $$ L_{combined} = 10 \log_{10} \left( 10^{\frac{L_1}{10}} + 10^{\frac{L_2}{10}} \right) $$ where $L_1$ is the sound pressure level of the shovel and $L_2$ is the sound pressure level of the dumper.

2. Substitute Known Values We know:

• Combined level $L_{combined} = 95 , \text{dB(A)}$
• Shovel level $L_1 = 90 , \text{dB(A)}$

Now substituting these values into the formula: $$ 95 = 10 \log_{10} \left( 10^{\frac{90}{10}} + 10^{\frac{L_2}{10}} \right) $$

3. Solve for $L_2$ First, isolate the logarithm: $$ 10^{\frac{95}{10}} = 10^{\frac{90}{10}} + 10^{\frac{L_2}{10}} $$

Calculating the left side: $$ 10^{9.5} = 31622.7766 $$ And the right side: $$ 10^{9} = 10000 $$

Now we can set up the equation: $$ 31622.7766 = 10000 + 10^{\frac{L_2}{10}} $$

Subtract 10000 from both sides: $$ 21622.7766 = 10^{\frac{L_2}{10}} $$

4. Convert Back to dB Now to solve for $L_2$: $$ \frac{L_2}{10} = \log_{10}(21622.7766) $$ Calculating the logarithm: $$ \log_{10}(21622.7766) \approx 4.3366 $$

Multiply by 10 to find $L_2$: $$ L_2 \approx 43.366 , \text{dB} $$

Finally, rounding to two decimal places gives: $$ L_2 \approx 43.37 , \text{dB} $$