A chain saw produces a spherical sound wave having a frequency of 214Hz in air at 358C (308.2K or 958F). At a distance of 600mm (23.62 in), the sound pressure level is 100 dB. Determine (a) the acoustic power level and (b) the rms acoustic particle velocity and velocity level at a distance of 600mm from the saw.

Given Information:

Frequency = 214 Hz

Temperature = 358° C = 308.2 K

Sound Pressure = p = 100 dB = 2 pascal

Distance = 600 mm = 0.60 m

Required Information:

(a) Acoustic Power Level = ?

(b) Acoustic Particle Velocity = ?

and Velocity level at a distance of 600 mm = ?

Acoustic Power Level = Lw = 106.44 dB

Acoustic Particle Velocity = v = 0.0106 m/s

Velocity level = 60.25 dB

Explanation:

(a) Acoustic Power Level

Acoustic Power = W = 4πr² I

Acoustic Intensity = I =  p ²/Z₀

Where Z₀ is the characteristic impedance of air Z₀ = 409.8 rayl

I =  p ²/Z₀ = (2)²/409.8 = 0.00976 W/m²

W = 4πr² I = 4*π(0.60)²*0.00976 = 0.0441 W

Acoustic Power Level = Lw = 10log(W/Wref)

Where Wref is Reference Acoustic Power Wref = 1x10⁻¹² W

Lw = 10log(W/Wref) = 10log(0.0441/1x10⁻¹²) = 106.44 dB

Lw = 106.44 dB

(b) Acoustic Particle Velocity

Acoustic Particle velocity = v =  p ²/Zs

Where Zs is specific acoustic impedance

Zs =  Z₀kr/(1 + k²r²)⁰°⁵

Where k = 2πf/c and c = 346.1 m/s is the speed of sound in air

k = 2π*214/346.1 = 3.885 per m

Zs =  409.8*3.885*0.60/(1 + (3.885)²(0.60)²)⁰°⁵

Zs = 376.6 rayl

v =  p ²/Zs = 2²/376.6 = 0.0106 m/s

v = 0.0106 m/s

Velocity level = 10log(v/vref) where vref = 10x10⁻⁹ m/s

Velocity level = 10log(0.0106/10x10⁻⁹) = 60.25 dB

Velocity level = 60.25 dB

gans

explanation:

vestigial organs (like your appendix) are organs that no longer serve a purpose.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.