Answer:
a)
The wave amplitude A = 0.075 m
Angular frequency = 12.57 rad/s
Period = 0.5 s
Wavelength = 6.0 m
Wave number = 1.0475 m⁻¹
b)
The wave function describing the wave = [tex]y(x,t) = 0.075 m \ cos \ [(1.0475 \ rad/m)x - (12.57 \ rad/s) t][/tex]
c) y= 0.0748875 + 0.9760 t
Explanation:
a)
The wave amplitude A = 0.075 m
Angular frequency (ω) = 2πf
= 2× π× 2
= 12.57 rad/s
Period [tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{2}\\\\T= 0.5 \ s[/tex]
Wavelength [tex]\lambda = \frac{v}{f}[/tex]
[tex]\lambda = \frac{12}{2} \\\\\lambda = 6.0 \ m[/tex]
Wave number [tex]k = \frac{\omega }{v}[/tex]
[tex]k = \frac{12.57}{12}\\\\k = 1.0475 \ m^{-1}[/tex]
b)
The wave equation describing the wave can be illustrated as:
[tex]y(x,t) = A \ cos \ (kx - \omega t)[/tex]
[tex]y(x,t) = 0.075 m \ cos \ [(1.0475 \ rad/m)x - (12.57 \ rad/s) t][/tex]
c)
The equation for the displacement as a function of time of the particle at and of a point at 3.00 m is as follows:
[tex]y= 0.075 \ cos \ [(1.0475 *3) - (12.57 ) t][/tex]
[tex]y= 0.075 *0.9985+ 0.9760 \ t\\\\y = 0.0748875 + 0.9760 t[/tex]
