Question

The sound from a trumpet radiates uniformly in all directions in [math]20^{\circ} C[/math] air. At a distance of 5.00 m from the trumpet the sound intensity level is 52.0 dB. The frequency is 587 Hz. (a) What is the pressure amplitude at this distance? (b) What is the displacement amplitude? (c) At what distance is the sound intensity level 30.0 dB?

Answer 1

Part a)

Pressure amplitude is

`P = 1.14 * 10^[-2} Pa`

Part b)

displacement amplitude is given as

`A = 7.49 * 10^(-9) m`

Part c)

Distance at the other intensity is

`r_2 = 63 m`

Step-by-step explanation:

Part a)

As we know that intensity of sound at 5 m distance is 52 dB

so we have

`52 = 10 Log((I)/(I_o))`

now we have

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

`I = 1.58 * 10^(-7) W/m^2`

Now we know that speed of sound at 20 degree C is given as

`v = 332 + 0.6 t`

`v = 332 + 0.6(20)`

`v = 344 m/s`

Now we know the relation of intensity of wave and pressure amplitude as

`I = (P^2)/(2\rho v)`

`1.58 * 10^[-7} = (P^2)/(2(1.2)(344))`

`P = 1.14 * 10^[-2} Pa`

Part b)

Now displacement amplitude with pressure amplitude is related as

`P = BAk`

now bulk modulus of air is given as

`B = 1.42 * 10^5 Pa`

also we know that

`\lambda = (v)/(f)`

`\lambda = (344)/(587) = 0.586 m`

now we have

`K = (2\pi)/(\lambda) = 10.72 m^(-1)`

so we have

`A = (1.14* 10^(-2))/((1.42 * 10^5)(10.72))`

`A = 7.49 * 10^(-9) m`

Part c)

As we know that intensity is inversely depends on the square of the distance

so we will have

`L_1 - L_2 = 10 Log((I_1)/(I_2))`

`L_1 - L_2 = 10 Log((r_2^2)/(r_1^2))`

`52 - 30 = 10 Log((r_2^2)/(5^2))`

`r_2 = 63 m`