Question

A source of sound operates 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m/s and the density of air is 1.2 kg /m³.
(a) What is the intensity at a distance of 6.0 m from the source?
(b) What will be the pressure amplitude at this point?
(c) What will be the displacement amplitude at this point?

Answer 1

Explanation ⇒

(a). Intensity = Power/Area of sphere drawn to that point taking the source of sound as an centre of sphere.

∴ Intensity = 20/[4π(6)²]

∴ Intensity = 44 × 10⁻³ W/m².

(b). We know that,

p₀² = 2ρVI, where ρ = density of air, V is the Speed of sound in air, and I is the intensity at an point.

∴ p₀² = 2 × 1.2 × 340 × 44 × 10⁻³

∴ p₀² = 36

∴ p₀ = 6 Pa.

(c). We know,

I =p₀(2π)s₀/2, where is frequency and s₀ is the displacement amplitude.

∴ s₀ = 2I/p₀(2π)

∴ s₀ = 2 × 44 × 10⁻³/6 × 2π × 2000

∴ s₀ = 1.2 × 10⁻⁶ m.

Hope it helps.

Answer 2

Answer:

a) 0.04423 W/m²

b) 6.0 N/m²

c) 1.17 × 10⁻⁶ m

Explanation:

Given:

Frequency of the source of the sound, f = 2.0 kHz = 2 × 10³ Hz

Power emitted, P = 20 W

Speed of the sound in the air, v = 340 m/s

Density of the air, ρ = 1.2 Kg/m³

Now,

a) The intensity I is given as :

I =$\frac{P}{\textup{Area of the sphere formed}}$

Area of the sphere formed = 4πr²

here, r is the distance from the source

therefore,

I = $\frac{20}{4\pi\times6^2}$

or

I = 0.04423 W/m²

or

I = 44.23 mW/m²

b) The pressure amplitude is given as:

P₀² = 2Iρv

on substituting the respective values, we get

P₀² = 2 × 0.04423 × 1.2 × 340

or

P₀² = 36.093

or

P₀ = 6.00 N/m²

c) Now,

The we have the relation

I = 2π²s²v²ρv

here,

s₀ is the displacement amplitude

on substituting the respective values, we get

0.04423 = 2 × π² × s² × 2000² × 1.2 × 340

or

s = 1.17 × 10⁻⁶ m