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
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.
To calculate the sound level in decibels (dB) of the given sound intensity, you correctly used the formula 10log(I/Io), where I is the sound intensity in watts per square meter (W/m^2), and Io is the reference sound intensity of 10^-12 W/m^2.
Now let's move on to part (b) of the question, where you need to determine the factor by which the amplitude of the sound wave would need to change to decrease the sound level by 8 dB.
To solve this question, we need to understand the relationship between sound level in dB and the sound intensity. The sound level is directly proportional to the logarithm of the sound intensity. Therefore, a decrease of 8 dB corresponds to dividing the sound intensity by a certain factor.
The formula to calculate the factor by which the sound intensity changes is as follows:
Factor = 10^(ΔL/10),
where ΔL represents the change in sound level in decibels. In this case, ΔL is -8 dB since we want to decrease the sound level by 8 dB.
To find the factor, we substitute ΔL = -8 dB into the formula:
Factor = 10^(-8/10),
Calculating this expression:
Factor = 10^(-0.8) = 0.1585.
So, the factor by which the amplitude of the sound wave would need to change to decrease the sound level by 8 dB is approximately 0.1585.
Please let me know if I can help you with anything else.