A highway carrying 5,000 vehicles per hour produces an overall noise level of ap

Question

A highway carrying 5,000 vehicles per hour produces an overall noise level of approximately 80 dB.

(a) Assuming that intensity scales linearly with the source of the noise, what is the increase in sound level if a highway is widened by a factor ×2 to permit twice as many cars on the road? A noise barrier can achieve a 5 dB noise level reduction, when it is tall enough to break the line-of-sight from the highway to a person’s home.

(b) What is the change in sound intensity due to this barrier?

(c) What is the equivalent distance to which a homeowner would have had to move to achieve the same level of reduction in noise as the barrier provides? After it breaks the line-of-sight, they estimate that a barrier can achieve approximately 1.5 decibels of additional noise level reduction for each additional meter of barrier height.

(d) What is the change in sound intensity due to this heightened barrier?

(e) What is the equivalent distance to which a homeowner would have had to move to achieve the same level of reduction in noise as the heightened barrier provides?

Explanation / Answer

The relation between the noise level and intensity I is

= 10log(I1/I0) ---- (1)

where I0 = a donstant refrerence intensity = 10-12W/m2

Intitally the noise level is 80 dB. So let the intitial intensity be I1.

So, 80dB = 10log(I1/I0)

where I0 = a donstant refrerence intensity = 10-12W/m2

a) When there are twice as many cars on the road, the intensity becomes 2I1 as the  intensity scales linearly with the source of the noise.

So new level of noise = 10log(2I1/I0) = 10log2 + 10log(I1/I0) = 3dB + 80dB = 83dB

So, the increase in sound level is = 83dB - 80dB = 3dB

b) After the cars are doubled the noise level is 83dB. But the barrier resuces it by 5dB. So new noise level is 78dB.

Let Intensity corresponding to 83dB be I.

then

10log(I/(10-12W/m2)) = 83 (using euqation (1))

or I = 1.99X10-4W/m2

Let Intensity corresponding to 78dB be I1.

Then 10log(I1/(10-12W/m2)) = 78

or I1 = 6.3X10-5W/m2

So change in intensity = 1.99X10-4W/m2 - 6.3X10-5W/m2 = 1.36X10-5W/m2

c)

Let at the intital position the distance d1 when thn intenisty is I = 1.99X10-4W/m2

Let he moves to distanc d2 so that intesity becomes I1 = 6.3X10-5W/m2

Intensity falls of proportional to the inverse of square of distance.

I/I1 = d22/d12

or d2/d1 = (1.99/0.63)

or d2 = 1.77d1

So he has to move 1.778 times the intital distance fom the source.

d) After it breaks the line-of-sight, a barrier can achieve approximately 1.5 decibels reduction.

Then the new level = 78dB-1.5dB = 76.5dB

Let the intensity be I

then 76.5 = 10log(I/I0), I0 = 10-12W/m2

or I = 4.46X10-5W/m2

Change in intesity = 6.3X10-5W/m2 - 4.46X10-5W/m2 = 1.84W/m2

This concludes the answers. Check the answer and let me know. If you need anymore clarification I will be happy to oblige...

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