A vacuum cleaner has a measured sound level of 75.7 dB. What is the intensity of this sound? Answer in units of W/m^2
B=75.7 dB
I=1x10^-2 W/m^2
B=10 log (I/I°)
I=I°10^3/10=(1x10^-2 W/m^2) 10^7.57
=3.71535x10^-5 W/m^2
To find the intensity of the sound, we can use the formula:
\[ I = 10^{\left(\frac{L}{10}\right)} \]
Where:
- \( I \) is the sound intensity in W/m²
- \( L \) is the sound level in dB
Given that the sound level is 75.7 dB, we can substitute this value into the formula:
\[ I = 10^{\left(\frac{75.7}{10}\right)} \]
Now, let's calculate the value of \( I \):
\[ I = 10^{\left(\frac{75.7}{10}\right)} \]
\[ I = 10^{7.57} \]
\[ I \approx 555467.6313 \]
Therefore, the intensity of this sound is approximately 555467.6313 W/m².
To find the intensity of the sound, we can use the formula:
Intensity = 10^((Sound Level - Sound Level Reference) / 10)
In this case, the sound level reference is commonly taken as 0 dB, which corresponds to the threshold of hearing. Therefore, we can substitute the values into the formula, with the sound level of the vacuum cleaner (measured) as 75.7 dB:
Intensity = 10^((75.7 dB - 0 dB) / 10)
We can simplify the equation:
Intensity = 10^(7.57)
Using a calculator or computational software, we can evaluate this expression:
Intensity ≈ 50118.72 W/m^2.
So, the intensity of the sound produced by the vacuum cleaner is approximately 50118.72 W/m^2.