Plane harmonic waves of frequency 500 Hz are produced in air with displacement amplitude of `10mum`. Given that density of air is `1.29(kg)/(m^3)` and speed of sound in air is `340(m)/(s)`. Then

Options

A. the pressure amplitude is `13.8(N)/(m^2)`
B. the energy density is `6.4xx10^-4(J)/(m^3)`
C. the energy flux is `0.22(J)/((m^2s))`
D. only (a) and (c ) are correct

Answer

A, B, C

Explanation

Pressure amplitude
`P=(2piaE)/(lamda)`
where `E` is the coefficient of elasticity. We have
`E=v^2rho{becausev=sqrt((E)/(rho))}`
`P_0=(2piav^2rho)/(lamda)=(2piav^2rho)/(((v)/(f)))`
`P_0=2piavfrho=13.8(N)/(m^2)`
Energy density is `2pi^2a^2f^2rho=6.4xx10^-4(J)/(m^3)`
Energy flux is `2pi^2a^2f^2rhov=0.22(J)/((m^2)/(s))`