The intensity of a 50-dB sound is reduced by one-half. What is the new intensity of the sound in dB?
I have no idea, but I would think it would be 25 dB.
50 dB sound intensity level equals 10-7 W/m2 = 0.0000001 W/m2
To find the new intensity of the sound in dB, we first need to understand that the decibel scale is logarithmic. The formula to calculate the change in sound intensity in decibels is given by:
ΔdB = 10 * log10(I2/I1)
where ΔdB is the change in decibels, I1 is the initial intensity, and I2 is the final intensity.
In this case, we are given that the initial sound intensity is 50 dB. Since we want to find the new intensity after it is reduced by one-half, we can express the final intensity as:
I2 = (1/2) * I1
Substituting this value into the formula, we get:
ΔdB = 10 * log10((1/2) * I1 / I1)
= 10 * log10(1/2)
= 10 * (-0.3010)
= -3.010
So, the change in decibels is -3.010 dB. To find the new intensity in dB, we subtract this change from the initial intensity:
New intensity in dB = 50 dB - 3.010 dB
= 46.990 dB
Therefore, the new intensity of the sound in dB is approximately 46.990 dB.