The sound intensity level at a concert increases from 85 dB to 94 dB
when the concert begins. By what factor has the sound intensity increased?
each 3 db means doubling
88 db has doubled
91 db has doubled again
94 db has doubled again.
Intensity has increased 2^3 times,or 8x
To find the factor by which the sound intensity has increased, we can use the formula for sound intensity level in decibels (dB):
L2 = L1 + 10 * log(I2/I1)
Where:
L2 = final sound intensity level (94 dB)
L1 = initial sound intensity level (85 dB)
I2 = final sound intensity
I1 = initial sound intensity
First, we need to isolate the ratio of sound intensities (I2/I1) using the equation:
L2 - L1 = 10 * log(I2/I1)
94 dB - 85 dB = 10 * log(I2/I1)
9 dB = 10 * log(I2/I1)
Next, divide both sides of the equation by 10:
0.9 = log(I2/I1)
Now, we need to convert the logarithmic equation into exponential form:
10^0.9 = I2/I1
Using a calculator, we find that 10^0.9 ≈ 7.943
So, the factor by which the sound intensity has increased is approximately 7.943. Therefore, the sound intensity has increased by a factor of about 7.943.