displacement, pressure amplitude & distance to sound level

Question

A certain source of sound radiates uniformly in all directions in air. At a distance of 5 m the sound level is 80 dB. The frequency is 440 Hz. (a)What is the displacement amplitude at this distance? (b) What is the pressure amplitude? (c) At what distance is the sound level 60 dB?

Answer 1

(a) What is the displacement amplitude at this distance?

We want to find the displacement amplitude Δs in terms of quantities we are either given or we can calculate fairly easily. You should be able to find this equation in your text or class notes :

I = 2 × π² × ρ × f² × v × (Δs)², where

I : Intensity in W/m²

ρ : density of the medium (air) in kg/m³

f : frequency of the wave in Hz or 1/s

v : speed of the wave in m/s

Δs : displacement amplitude in m

We can rearrange that equation to solve for Δs:

Δs = √ ( I ⁄ [ 2 × π² × ρ × f² × v ] )

We can calculate the intensity I from the "intensity level" (also know as the "sound level") L, which is given as 80 dB at a distance of 5 m from the source of the sound.

I and L are related by :

I = I₀ x 10^{( L/10)}

I₀ is the reference intensity for sound, and correspond to 0 dB. It is roughly the minimum sound intensity that humans can typically hear, and is 10^-12 W/m².

Using that expression, we can rewrite Δs:

Δs = √ (( I₀ × 10^(L/10) ) /(2 × Π ² × ρ × f² × v ))

... which gives Δs in terms of only known values:

I₀ = 1 × 10⁻¹² W/m²

L = 80 dB

ρ = 1.29 kg/m³

f = 440 Hz

v = 343 m/s (at 20 degrees C; roughly room temperature)

Plugging those values into the equation above gives:

Δs = 2.476 × 10^-7 m

(b) What is the pressure amplitude?

Starting from the relation between the pressure amplitude ΔP and the displacement amplitude Δs:

∆P = 2πρfv∆s

... we know the values of all the quantities on the right hand side:

ρ = 1.29 kg/m³

f = 440 Hz

v = 343 m/s (at 20 degrees C)

Δs = 2.476 × 10^-7 m

Replacing those values in the expression for ΔP gives

ΔP = 0.3029 kg/(m × s²)

Since 1 kg/(m × s²) = 1 Pa, ΔP = 0.3029 Pa

(c) At what distance is the sound level 60 dB?

First we find the intensity that corresponds to 60 db. From part 'a', the intensity I and the intensity level L are related by :

I₆₀ = I₀ × 10^(L/10)

Using the reference intensity I₀ = 10⁻¹² W/m² , and given L = 60 dB:

I₆₀ = 1.0^-6 W/m²

To find the distance r where this sound level occurs, we can use the fact that the intensity varies as the inverse square of the distance from the source:

I ∝ 1/r²

Using our results from part 'a' for the distance of 5 m, we can take the ratio of that intensity (call it I₈₀ = 10^-4 W/m²) to our intensity here (I₆₀) corresponding to 60 dB to calculate our unknown distance r₆₀:

I₈₀/I₆₀ = (r₆₀/r₈₀)²

So r₆₀ = r₈₀ ×√(I₈₀/I₆₀)

We know:

r₈₀ = 5 m

I₈₀ = 10^-4 W/m²

I₆₀ = 10^-6 W/m²

...so:

r₆₀ = 50 m