Since the sond level is less than the
reference(1Watt), the answer will be negative. If the sound level was 1Watt,
the answer would be 0db. If the sound
level was greater than 1 Watt, the answer would be positive.
a. 10Log(P2/P1) = 10Log(6.3*10^-4W/1W) = 10 * (-3.2) = -32db.
b. -32db - 8db = -40db = The new power
ratio in decibels.
10Log(P2/1W) = -40db,
Divide both sides by 10:
Log(P2/1) = -4.0,
P2/1 = 10^-4,
P2 = 10^-4 Watts = The new power.
6.3*10^-4W / 10^-4W = 6.3 = Factor
by which the power was decreased.
Since the power is proportional to the
square of the voltage,the voltage will be reduced by a factor of:
sqrt(6.3) = 2.51.
Factory Worker A is exposed to a sound of a constant intensity of 6.30 × 10–4 W m–2 at a distance of 1.00 m from a noisy machine. The frequency of the sound is 4.00 kHz.
(a) What is the sound level in decibels (dB) of this sound?
(b) By what factor would the amplitude of this sound wave need to change to decrease the sound level by 8 dB?
I think the first answer a) is 10log(6.3*10^-4/10^-12)=88dB but i'm unsure how to answer b
1 answer
1 answer