Two sounds have measured intensities of I1 = 110 W/m2 and I2 = 330.00 W/m2. By how many decibels is the level of sound 1 lower than that of sound 2? Answer in dB.
db = 10*Log(I1/I2) = 10*Log(110/330) =
-4.77
I1 = 4.77 db below I2.
To calculate the difference in decibels between two sound levels, we can use the formula:
Difference in decibels (dB) = 10 * log10 (I2 / I1)
where I1 and I2 represent the intensities of sound 1 and sound 2, respectively.
Given the intensities provided:
I1 = 110 W/m2
I2 = 330 W/m2
Substituting these values into the formula, we get:
Difference in decibels (dB) = 10 * log10 (330 / 110)
To calculate this, follow these steps:
1. Divide the intensity of sound 2 by the intensity of sound 1: (330 / 110) = 3.
2. Take the logarithm base 10 of the result: log10 (3) ≈ 0.477.
3. Multiply the logarithm by 10: 0.477 * 10 = 4.77 dB (rounded to two decimal places).
Therefore, the level of sound 1 is approximately 4.77 dB lower than that of sound 2.