Question

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pressure amplitude, (b) the frequency, (c) the wavelength, and (d) the speed of the wave .

Answer 1

a)
`P_m=1.78\ Pa`

b)
`f=79.5775\ Hz`

c)
`\lambda=7.076\ m`

d)
`v=563.06\ m.s^(-1)`

Step-by-step explanation:

Given equation of pressure variation:

`\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^(-1))x-(500\ s^(-1))t]`

We have the standard equation of periodic oscillations:

`\Delta P=P_m\ sin\ (kx-\omega.t)`

By comparing, we deduce:

(a)

amplitude:

`P_m=1.78\ Pa`

(b)

angular frequency:

`\omega=2\pi.f`

`2\pi.f=500`

∴Frequency of oscillations:

`f=(500)/(2\pi)`

`f=79.5775\ Hz`

(c)

wavelength is given by:

`\lambda=(2\pi)/(k)`

`\lambda=(2\pi)/(0.888)`

`\lambda=7.076\ m`

(d)

Speed of the wave is gives by:

`v=(\omega)/(k)`

`v=(500)/(0.888)`

`v=563.06\ m.s^(-1)`