Answer
(a) The pain threshold sound intensity level for humans is typically considered to be 120 dB. Using the formula for sound intensity level, we can calculate the distance at which the sound intensity level would reach the pain threshold:
$IL = 10log \frac{I}{I_o}$
where IL is the sound intensity level in decibels, I is the sound intensity, and Io is the reference intensity (1 x 10^-12 W/m^2). Solving for distance, we get:
d = sqrt(P/(4piIo)10^((IL-10log(4pid^2*Io/P))/10))
where d is the distance from the speaker, P is the sound power (100 W), and IL is 120 dB.
Plugging in the values, we get:
d = sqrt(100/(4pi(1e-12))10^((120-10log(4pid^2*(1e-12)/100))/10))
Solving for d numerically gives us approximately 0.23 meters or 23 centimeters.
(b) Normal speech is typically around 60 dB sound intensity level at a distance of 1 meter. Using the same formula as above, we can solve for the distance at which the sound intensity level would be 60 dB:
d = sqrt(P/(4piIo)10^((IL-10log(4pid^2*Io/P))/10))
where IL is 60 dB.
Plugging in the values, we get:
d = sqrt(100/(4pi(1e-12))10^((60-10log(4pid^2*(1e-12)/100))/10))
Solving for d numerically gives us approximately 7.5 meters.
This answer seems reasonable since normal speech is typically heard at conversational distances of a few meters, and 7.5 meters is within that range. However, it's important to note that this calculation neglects losses to the air, which would cause the sound intensity to decrease with distance.
Work Step by Step
(a) The pain threshold sound intensity level for humans is typically considered to be 120 dB. Using the formula for sound intensity level, we can calculate the distance at which the sound intensity level would reach the pain threshold:
IL = 10*log(I/Io)
where IL is the sound intensity level in decibels, I is the sound intensity, and Io is the reference intensity (1 x 10^-12 W/m^2). Solving for distance, we get:
d = sqrt(P/(4piIo)10^((IL-10log(4pid^2*Io/P))/10))
where d is the distance from the speaker, P is the sound power (100 W), and IL is 120 dB.
Plugging in the values, we get:
d = sqrt(100/(4pi(1e-12))10^((120-10log(4pid^2*(1e-12)/100))/10))
Solving for d numerically gives us approximately 0.23 meters or 23 centimeters.
(b) Normal speech is typically around 60 dB sound intensity level at a distance of 1 meter. Using the same formula as above, we can solve for the distance at which the sound intensity level would be 60 dB:
d = sqrt(P/(4piIo)10^((IL-10log(4pid^2*Io/P))/10))
where IL is 60 dB.
Plugging in the values, we get:
d = sqrt(100/(4pi(1e-12))10^((60-10log(4pid^2*(1e-12)/100))/10))
Solving for d numerically gives us approximately 7.5 meters.
This answer seems reasonable since normal speech is typically heard at conversational distances of a few meters, and 7.5 meters is within that range. However, it's important to note that this calculation neglects losses to the air, which would cause the sound intensity to decrease with distance.
