Question

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b) What is the pressure amplitude of the sound?

Answer 1

Part a)

`I = 3.16 * 10^(-5) W/m^2`

Part b)

`P_o = 0.162 Pa`

Step-by-step explanation:

Part a)

Level of sound = 75 dB

now we know that

`L = 10 Log(I)/(I_0)`

here we know that

`I_0 = 10^(-12) W/m^2`

now we have

`75 = 10 Log((I)/(10^(-12)))`

`I = 3.16 * 10^(-5) W/m^2`

Part b)

Intensity of sound wave is given as

`I = (1)/(2)\rho A^2\omega^2 c`

here we know that

`A = (P_o)/(Bk)`

so we have

`I = (1)/(2)\rho((P_o)/(Bk))^2\omega^2 c`

`I = (1)/(2)\rho P_o^2 (c^3)/(B^2)`

now we know

`\rho = 1.2 kg/m^3`

`c = 340 m/s`

`B = 1.4 * 10^5 Pa`

now we have

`3.16 * 10^(-5) = (1)/(2)(1.2)P_o^2(340^3)/((1.4* 10^5)^2)`

`P_o = 0.162 Pa`