The sound from a trumpet radiates uniformly in all directions in air with a temp

Question

The sound from a trumpet radiates uniformly in all directions in air with a temperature of 20C. At a distance of 4.98 m from the trumpet the sound intensity level is 55.0 dB . The frequency is 583 Hz .

What is the pressure amplitude at this distance? Use 344 m/s for the speed of sound in air, 1.20 kg/m3 for the density of air, and 1.00×1012 W/m2 for the reference intensity.

What is the displacement amplitude? Use 1.42×105 Pa for the adiabatic bulk modulus of air.

At what distance is the sound intensity level 35.0 dB ?

Explanation / Answer

here,
temperature, T = 20 C
Sound intensity Level = 55dB
Frequency, f = 583 Hz

Sound intensity level, dB = 10*log(I/Io)
Where,
Io = is reference intensity
I = Intensity in Watt/m^2
log is bas 10 logarithms

rewriting above equation of intensity of sound :

I/Io = 10^(dB/10)
I/Io = 10^(55/10)
I = 316227.766 * 10^-12
I = 3.162*10^-7 watts/m^2

pressure ampitude is given as :

P = sqrt(2*rho*I*c)
P = sqrt(2*1.20*3.162*10^-7*344)
P = 0.016 pa

Part b:
Pressure amplitude = B*(2*pi*f)*A/C
Where,
C = speed of sound
f = Frequency
B = Bulk modulus

rewriting eqn for displacement amplitude, A

A = P*C /B*(2*pif)
A = 0.016*344 /(1.42*10^5*(2*3.14*583)
A = 1.059*10^-8 m

Part C:
I1 = 55 dB
I2 = 35 dB
r1 = distance at which I1
r2 = distance at which I2

as I1*r1^2 = I2*r2^2

rewriting eqn for r2

r2 = sqrt(I1*r1^2 / I2)
r2 = r1*sqrt(55/35)
r2 = 1.254*r1
r2 = 1.254 * 4.98
r2 = 6.245 m

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