Answer:
a) [tex]P_m=1.78\ Pa[/tex]
b) [tex]f=79.5775\ Hz[/tex]
c) [tex]\lambda=7.076\ m[/tex]
d) [tex]v=563.06\ m.s^{-1}[/tex]
Explanation:
Given equation of pressure variation:
[tex]\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t][/tex]
We have the standard equation of periodic oscillations:
[tex]\Delta P=P_m\ sin\ (kx-\omega.t)[/tex]
By comparing, we deduce:
(a)
amplitude:
[tex]P_m=1.78\ Pa[/tex]
(b)
angular frequency:
[tex]\omega=2\pi.f[/tex]
[tex]2\pi.f=500[/tex]
∴Frequency of oscillations:
[tex]f=\frac{500}{2\pi}[/tex]
[tex]f=79.5775\ Hz[/tex]
(c)
wavelength is given by:
[tex]\lambda=\frac{2\pi}{k}[/tex]
[tex]\lambda=\frac{2\pi}{0.888}[/tex]
[tex]\lambda=7.076\ m[/tex]
(d)
Speed of the wave is gives by:
[tex]v=\frac{\omega}{k}[/tex]
[tex]v=\frac{500}{0.888}[/tex]
[tex]v=563.06\ m.s^{-1}[/tex]
