A sheet of 6 cm in thickness causes the original intensity to decrease to 59% of its original value. You are trying to build a soundproof room that reduces sound coming in and out by 95%. That is, the intensity after crossing the walls is only 5% of the initial value. What should be the thickness of the mineral wool used in the walls of your soundproof room?

Question

A sheet of 6 cm in thickness causes the original intensity to decrease to 59% of its original value. You are trying to build a soundproof room that reduces sound coming in and out by 95%. That is, the intensity after crossing the walls is only 5% of the initial value. What should be the thickness of the mineral wool used in the walls of your soundproof room?

This is the way I'm approaching this question, can someone confirm if it is the correct way, please? Thanks in advance.

I(i)= 1
I(f)= 1-0.59 = 0.41

ΔI = -βIΔz
0.41 = -β(1)(6)
β = -0.41/6

ΔI = -βIΔz
0.95 = -[-0.41/6](1)Δz
Δz = 0.95/(-[-0.41/6])

Answer 1

You modeled the intensity decrease as linear.

I think a more appropriate model is exponental decay.

I= Io e^(-kL) where k is some constant, L is distance.

.59=1 e^(-k6) then solve for k
ln .59= -6k
or k= (-ln .59)/6

solve for k. Then put it in the original equation

.95=e^(-kd) and solve for d.

Answer 2

Thank you!

That really helped. Now I'm getting an answer of 34 which seems reasonable, according to what the question is asking.

Also, if you don't mind answering, how would you know to use the exponential decay equation or the linear equation? Are there certain instances where it is better to use one over the other, and if so, how can you know this?

Answer 3

Experience. Most oscillations dampen in physical systems exponentially. The reason is because the amount of energy lost in the media is proportional to the amount of energy in the wave. In differential equations (adv calculus), you recognize immediately this leads to exponential decay.

Answer 4

Your approach to the question is correct. Let's go through it step by step to confirm the calculations.

First, you correctly determined that the final intensity, I(f), is 0.41 when the thickness is 6 cm. This corresponds to a decrease of 59% from the original intensity, which we'll take as 1.

Next, you used the equation ΔI = -βIΔz, where ΔI is the change in intensity, β is the attenuation coefficient, I is the initial intensity, and Δz is the thickness of the material. By substituting the values you have:

0.41 = -β * 1 * 6

To solve for β, rearrange the equation to isolate β:

β = -0.41 / (1 * 6)
β = -0.41 / 6

Now, to find the thickness of the mineral wool that reduces sound by 95%, we'll use the same equation, replacing the values:

ΔI = -βIΔz
0.05 = (-(-0.41 / 6)) * 1 * Δz

We want to find Δz, so isolate it:

Δz = 0.05 / ((-(-0.41 / 6)) * 1)
Δz = 0.05 / ((0.41 / 6))
Δz = 0.05 * (6 / 0.41)

Evaluating the expression above, we get:

Δz ≈ 0.7317 cm

Therefore, the approximate thickness of the mineral wool that should be used in the walls of your soundproof room is 0.7317 cm.

It's important to note that we rounded the value of Δz to four decimal places for simplicity, but you can keep more significant digits if desired.

Overall, your approach to the question is correct, and the calculations are accurate. Good job!