Sound radiation from a pulsating sphere

[mk55]

Homework Statement

Consider a pulsating sphere of radius 0.15m pulsating at a frequency of 2000Hz and surface velocity v[itex]_{}0[/itex]=0.07m/s. What is the distance r, where the measured acoustic pressure is 5x10[itex]^{}-2[/itex].
speed of sound c=343m/s
mass density of air ρ[itex]_{}0[/itex]=1.25kg/m3

Homework Equations

i know I am supposed to use these equations:
φ=(A/r)e^(ikr-iωt)
p'=-ρ[itex]_{}0[/itex](dφ/dt)
v=dφ/dr
v=v[itex]_{}0[/itex]e^(-iωt)

The Attempt at a Solution

I have differantiated the velocity potential function (φ) and substituted in the formula for p'. i also substituted the value of p' which is 5x10^-2 but after that I am stuck because I am not sure how to work out t.

Any help would be appreciated

 

[KataKoniK]

.

Thank you for your post. It seems like you are on the right track with using the equations provided to solve for the distance r where the measured acoustic pressure is 5x10^-2. Here are the steps I would recommend taking to find the solution:

1. Start by setting up the equation for the velocity potential, φ=(A/r)e^(ikr-iωt), where A is the amplitude, k is the wave number, and ω is the angular frequency.

2. Use the given values for the radius, frequency, and surface velocity to solve for the amplitude A.

3. Next, differentiate the velocity potential with respect to time to find the particle velocity, v=dφ/dt.

4. Substitute the given value for the surface velocity v_0=0.07m/s and the particle velocity v into the equation v=v_0e^(-iωt) to solve for ωt.

5. Use the given value for the speed of sound c=343m/s to solve for the wave number k.

6. Now, substitute the values for A, k, and ωt into the velocity potential equation and solve for r. This will give you the distance at which the measured acoustic pressure is 5x10^-2.

I hope this helps you find the solution you are looking for. it is important to carefully consider all the given information and use the appropriate equations to find a solution. Keep up the good work!