Two sounds have measured intensities of I1 = 100 W/m2 and I2 = 300 W/m2. By how many decibels is the level of sound 1 lower than that of sound 2?
To calculate the difference in decibels between two sound levels, we can use the formula:
dB = 10 * log10(I1/I2),
where dB is the difference in decibels and I1 and I2 are the intensities of sound 1 and sound 2, respectively.
Given:
I1 = 100 W/m2
I2 = 300 W/m2
Substituting these values into the formula, we have:
dB = 10 * log10(100/300)
Now, let's calculate:
dB = 10 * log10(1/3)
Since log10(1/3) is approximately -0.52, we can substitute this value back into the equation:
dB = 10 * (-0.52)
Calculating further gives us:
dB ≈ -5.2
Therefore, the level of sound 1 is lower than the level of sound 2 by approximately 5.2 decibels.
To calculate the difference in decibels (dB) between two sound levels, you can use the following formula:
ΔL = 10 * log10(I2/I1)
Where ΔL is the difference in decibels, I1 is the intensity of sound 1, and I2 is the intensity of sound 2.
Given that I1 = 100 W/m^2 and I2 = 300 W/m^2, we can substitute these values into the formula:
ΔL = 10 * log10(300/100)
First, divide the intensity of sound 2 by the intensity of sound 1:
ΔL = 10 * log10(3)
Then, calculate the logarithm of the result:
ΔL = 10 * 0.477
Finally, multiply the result by 10:
ΔL ≈ 4.77 dB
Therefore, the level of sound 1 is approximately 4.77 decibels lower than that of sound 2.