Question

Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardrums of the first student is 0.58 W/m^2, while at the eardrums of the second student the sound intensity is 1.18 times greater.

Required:
a. What is the ratio of the diameter of the first student's eardrum to that of the second student?
b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

Answer 1

Answer:

a. d₁/d₂ = 1.09 b. 0.054 mW

Explanation:

a. What is the ratio of the diameter of the first student's eardrum to that of the second student?

We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So, I ∝ I/d²

I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.

Given that I₂ = 1.18I₁

I₂/I₁ = 1.18

Since I₁/I₂ = d₂²/d₁²

√(I₁/I₂) = d₂/d₁

d₁/d₂ = √(I₂/I₁)

d₁/d₂ = √1.18

d₁/d₂ = 1.09

So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09

b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

We know intensity, I = P/A where P = acoustic power and A = area = πd²/4

Now, P = IA

= I₂A₂

= I₂πd₂²/4

= 1.18I₁πd₂²/4

Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

So, P = 1.18I₁πd₂²/4

= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

= 0.691244π × 10⁻⁴ W/4 =

2.172 × 10⁻⁴ W/4

= 0.543 × 10⁻⁴ W

= 0.0543 × 10⁻³ W

= 0.0543 mW

≅ 0.054 mW

Answer 2

(a) The ratio of the diameter of the first student's eardrum to that of the second student is 1.09.

(b) The acoustic power, in microwatts, striking each of his (and the other student's) eardrums is of 0.054 mW.

Given data:

The sound intensity of first student is, $I_{1}= 0.58 \;\rm W/m^{2}$.

And sound intensity of second student is, $I_{2} = 1.18 \times I_{1}$.

The diameter of second eardrum is, $d_{2} = 1.01 \;\rm cm=1.01 \times 10^{-2} \;\rm m$

(a)

The power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So,

I ∝ I/d²

I₁/I₂ = d₂²/d₁²

Here

I₁  is the intensity at eardrum of first student.

d₁ is the diameter of first student's eardrum.

I₂ is the  intensity at eardrum of second student.

d₂ is the diameter of second student's eardrum.

Since,  I₂ = 1.18I₁

I₂/I₁ = 1.18

Also,  I₁/I₂ = d₂²/d₁²

=√(I₁/I₂) = d₂/d₁

=d₁/d₂ = √(I₂/I₁)

=d₁/d₂ = √1.18

d₁/d₂ = 1.09

Thus, we can conclude that  the ratio of the diameter of the first student's eardrum to that of the second student is 1.09.

(b)

We know that the expression for the intensity of sound is,

I = P/A

P = IA

Here,

P is the  acoustic power and A is the area. (A = πd²/4)

P = I₂A₂

P= I₂πd₂²/4

P= 1.18I₁πd₂²/4

Since,

I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

Substituting the values as,

P = (1.18I₁πd₂²) /4

P = 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

P = (0.691244π × 10⁻⁴ W) /4  

P = (2.172 × 10⁻⁴ W) /4

P = 0.543 × 10⁻⁴ W

P  ≅ 0.054 mW

Thus, we can conclude that the acoustic power, in microwatts, striking each of his (and the other student's) eardrums is of 0.054 mW.

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