An audio speaker at a concert emits sound uniformly in all directions at a rate
of 100 W.
(i) Calculate the sound intensity experienced by a listener at a distance
of 8 m from the speaker.
The listener moves back from the speaker to protect her hearing.
(ii) At what distance
from the speaker is the sound intensity level reduced by 3 dB?
I = P/A = P/4πr² = 100/4π(8)² = 0.12 W•m⁻²
If the sound intensity level is halved to a value of I = 0.06 W•m⁻²,
the sound intensity level will decrease by 3 dB.
I = P/4πr² = 100/4π(x) ² =
=0.06 W•m⁻²
x = 11.5 m from the speaker.
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To calculate the sound intensity experienced by the listener at a distance of 8m from the speaker, we can use the inverse square law which states that the intensity of sound decreases inversely proportional to the square of the distance from the source.
(i) Calculate the sound intensity experienced by the listener at a distance of 8m from the speaker:
The formula for sound intensity is:
I = P / (4πr^2)
where:
I = sound intensity
P = power emitted by the speaker
r = distance from the speaker
Given:
P = 100 W
r = 8 m
Using the formula, we can calculate the sound intensity:
I = 100 / (4π(8)^2)
I = 100 / (256π) ≈ 0.123 W/m^2
Therefore, the sound intensity experienced by the listener at a distance of 8m from the speaker is approximately 0.123 W/m^2.
(ii) At what distance from the speaker is the sound intensity level reduced by 3 dB:
To calculate the distance from the speaker at which the sound intensity level is reduced by 3 dB, we need to use the formula:
L2 = L1 - 10log(I2/I1)
where:
L1 = initial sound intensity level
L2 = reduced sound intensity level
I1 = initial sound intensity
I2 = reduced sound intensity
Given:
L1 = 0 dB (reference level)
L2 = -3 dB (reduced by 3 dB)
Using the formula, we can calculate the reduced sound intensity:
L2 = L1 - 10log(I2/I1)
-3 = 0 - 10log(I2/I1)
-3/10 = log(I2/I1)
10^(-3/10) = I2/I1
0.501 = I2/I1
So, when the sound intensity level is reduced by 3 dB, the ratio of the reduced sound intensity to the initial sound intensity is 0.501.
To find the distance, we can use the inverse square law:
I2/I1 = (r2/r1)^2
where:
r1 = initial distance
r2 = reduced distance
Substituting the values, we get:
0.501 = (r2/8)^2
Taking the square root, we have:
r2/8 = √(0.501)
r2 = 8 * √(0.501)
r2 ≈ 7.08 m
Therefore, at a distance of approximately 7.08 m from the speaker, the sound intensity level is reduced by 3 dB.
To solve this question, we need to use the formulas related to sound intensity and the sound level.
(i) Sound intensity, I, is calculated using the formula:
I = P / A
where I is the sound intensity, P is the power emitted by the speaker, and A is the surface area of the sphere in which the sound is spreading.
Given that the power emitted by the speaker is 100 W, we can calculate the sound intensity at a distance of 8 m from the speaker.
First, we need to calculate the surface area of the sphere by using the formula:
A = 4πr^2
where A is the surface area and r is the distance from the speaker.
Plugging in the values, we get:
A = 4π(8^2)
= 256π
Now we can calculate the sound intensity (I) using the formula mentioned above:
I = 100 W / (256π)
≈ 0.123 W/m^2
Therefore, the sound intensity experienced by the listener at a distance of 8 m from the speaker is approximately 0.123 W/m^2.
(ii) The sound level, measured in decibels (dB), is given by the formula:
L = 10 log (I/I₀)
where L is the sound level, I is the sound intensity, and I₀ is the reference sound intensity.
Given that we need to find the distance from the speaker where the sound intensity level is reduced by 3 dB, we can rearrange the formula above to solve for I.
By substituting the sound level difference of 3 dB, we get:
3 = 10 log (I/I₀)
Dividing both sides by 10 and exponentiating with base 10, we get:
10^(3/10) = I / I₀
Simplifying, we can define a ratio:
I / I₀ = 10^(3/10)
Now, we can calculate the sound intensity reduction ratio:
I / I₀ = 10^(3/10)
I / 0.123 W/m^2 = 10^(3/10)
Solving for I, we get:
I = 0.123 * 10^(3/10) ≈ 0.368 W/m^2
To find the distance at which the sound intensity is reduced by 3 dB, we can rearrange the formula for sound intensity:
I = P / A
Again using the formula for the surface area of the sphere, we can write:
I = P / (4πr^2)
Rearranging for r, we have:
r = sqrt(P / (4πI))
Plugging in the values, we get:
r = sqrt(100 / (4π*0.368)) ≈ 4.59 m
Therefore, the distance from the speaker at which the sound intensity level is reduced by 3 dB is approximately 4.59 m.