At a recent rock concert, a dB meter registered 118 dB when placed 1.60 m in front of a loudspeaker on the stage.
(a) What was the power output of the speaker, assuming a hemispherical radiation of the sound and neglecting absorption in the air?
Answer in W
(b) How far away would the intensity level be a somewhat reasonable 85 dB?
Answer in m
To find the power output of the speaker, we need to use the relationship between sound intensity and power.
(a) The intensity of sound at a distance from the source is given by the formula:
I = P/A
Where I is the intensity of sound, P is the power output of the speaker, and A is the surface area of the hemisphere created by the sound waves.
Given that the intensity level is 118 dB at a distance of 1.60 m from the speaker, we need to convert this to the intensity in watts per square meter. The equation for converting the intensity level in decibels to the intensity in watts per square meter is:
I (w/m^2) = 10^((IL - 12)/10)
Where IL is the intensity level.
Substituting the given values:
I (w/m^2) = 10^((118 - 12)/10)
I (w/m^2) = 10^(106/10)
I (w/m^2) = 10^10.6
Using a calculator, we find that I ≈ 398.11 W/m^2.
Next, we need to find the surface area of the hemisphere. The surface area of a hemisphere is given by the formula:
A = 2πr^2
Where r is the distance from the speaker to the dB meter, which is 1.60 m in this case.
A = 2π(1.60)^2
A ≈ 16π
Substituting the values into the formula for intensity:
398.11 = P / (16π)
To find P, we need to multiply both sides of the equation by 16π:
398.11 * 16π = P
Using a calculator, we find that P ≈ 19956 W.
Therefore, the power output of the speaker is approximately 19956 W.
(b) To find the distance at which the intensity level is a reasonable 85 dB, we need to use the inverse square law for sound intensity:
IL1 - IL2 = 20log(r1/r2)
Where IL1 is the initial intensity level, IL2 is the final intensity level, r1 is the initial distance, and r2 is the final distance.
Substituting the given values:
85 - 118 = 20log(r1/r2)
-33 = 20log(r1/r2)
Now we can rearrange the equation to solve for the ratio of distances (r1/r2):
(r1/r2) = 10^((-33)/20)
(r1/r2) = 10^(-1.65)
Next, we solve for the final distance (r2). Let r1 be 1.60 m:
(r1/ r2) = (1.60 / r2) = 10^(-1.65)
To isolate r2, we divide both sides of the equation by 1.60:
(1.60 / r2) = 10^(-1.65)
r2 = 1.60 / 10^(-1.65)
Using a calculator, we find that r2 ≈ 7.34 m.
Therefore, the distance at which the intensity level is a somewhat reasonable 85 dB is approximately 7.34 m.
To solve this problem, we will use the formulas related to sound intensity and the inverse square law.
(a) To find the power output of the speaker, we can use the formula:
Sound intensity (I) = Power (P) / (4πr^2)
where r is the distance from the speaker.
Given that the sound level is 118 dB and the distance from the speaker is 1.60 m, we need to convert the sound level to sound intensity by using the formula:
I = 10^((L - 10log(r)) / 10),
where L is the sound level in dB, and log is the logarithm to the base 10.
Using these values, we can solve for the power output:
Step 1: Convert the sound level to sound intensity:
I = 10^((118 - 10log(1.60)) / 10)
I = 10^((118 - 10 * 0.2041) / 10)
I = 10^(115.7959 / 10)
I = 10^11.5796
I = 415680.0119 W/m^2
Step 2: Calculate the power output:
Power output (P) = I * 4πr^2
P = 415680.0119 * 4π * (1.60)^2
P ≈ 10^6.642 W
P ≈ 4102308.36 W
Therefore, the power output of the speaker is approximately 4102308.36 W.
(b) Now, let's find the distance at which the intensity level would be around 85 dB.
Using the formula for sound intensity:
I = 10^((L - 10log(r)) / 10)
We can rearrange the formula to solve for the distance (r):
r = 10^((L - 10log(I)) / 20)
Plugging in the values:
r = 10^((85 - 10log(415680.0119)) / 20)
r = 10^((85 - 10 * 5.6171) / 20)
r = 10^(85 - 56.171) / 20)
r = 10^2.2079
r = 168.336 m
Therefore, the distance at which the intensity level would be approximately 85 dB is approximately 168.336 m.