Question

The sound from a trumpet radiates uniformly in all directions in air with a temperature of 20?C. At a distance of 4.92 m from the trumpet the sound intensity level is 51.0 dB . The frequency is 562 Hz .

Part A

What is the pressure amplitude at this distance?

Use 344 m/s for the speed of sound in air, 1.20 kg/m^3 for the density of air, and 1.00×10^?12 W/m^2 for the reference intensity.

Part B

What is the displacement amplitude?

Use 1.42×10⁵ Pa for the adiabatic bulk modulus of air.

Part C

At what distance is the sound intensity level 34.0 dB ?

Answer 1

Given that :

frequency, f = 562 Hz

sound intensity level, = 51 dB

Part-A : using an equation, we have

= 10 log (I / I₀)

sound intensity is given by, I = I₀ (10)/10

where, I₀ = reference intensity = 10^-12 W/m²

then, we get

I = (10^-12 W/m²) (10)^5.1 =  1.25 * 10^-7 W/m²

The pressure amplitude at this distance which is given as -

p_max = 2 I v 2 (1.25 * 10^-7 W/m²) (1.2 kg/m³) (344 m/s)

p_max = 0.01015 Pa

Part-B : The displacement amplitude which will be given as -

using an equation, we have

p_max = B k A

where, B = adiabetic bulk modulus of air = 1.42 x 10⁵ Pa

k = wave number = 2f / v = 2 (3.14) (562 Hz) / (344 m/s)

then, we get

(0.01015 Pa) = (1.42 x 10⁵ Pa) (10.25 m^-1) A

A = (0.01015 Pa) / (14.5 * 10^5 Pa.m^-1)

A = 7* 10^-9 m

A = 7 x 10^-9 m

Part-C : At a distance, the sound intensity level 34 dB which is given as -

using an equation, we have

I₁ r₁² = I₂ r₂²

r₂ = r₁ (I₁ / I₂)^0.5 = r₁ [10^(₁ - ₂)/10]^0.5

r₂ = r₁ 10^(₁ - ₂)/20

r₂ = (4.92 m) 10^0.85

r₂ = 34.83 m