Answer :
Explanation:
The displacement equation is given by :
(a) [tex]y=A\ sin(\omega t\pm kx)[/tex]
Since, [tex]\omega=2\pi f[/tex]
[tex]\lambda=\dfrac{2\pi}{k}[/tex]
[tex]y=A\ sin(2\pi f t\pm kx)[/tex]
[tex]y=A\ sin(\dfrac{2\pi vt}{2\pi /k}\pm kx)[/tex]
[tex]y=A\ sin(kvt+kx)[/tex]
[tex]y=A\ sink(vt+x)[/tex]
The above equation is in terms of k and v.
(b) [tex]y=A\ sin(\omega t\pm kx)[/tex]
[tex]y=A\ sin(2\pi \dfrac{v}{\lambda}t\pm\dfrac{2\pi}{\lambda}x)[/tex]
[tex]y=Asin\dfrac{2\pi}{\lambda}(vt\pm x)[/tex]
(c) [tex]y=A\ sin(\omega t\pm kx)[/tex]
[tex]y=A\ sin(2\pi f t\pm \dfrac{2\pi}{\lambda} x)[/tex]
[tex]y=A\ sin\ 2\pi (ft\pm \dfrac{x}{\lambda})[/tex]
(d) [tex]y=A\ sin(\omega t\pm kx)[/tex]
[tex]y=A\ sin(2\pi ft+\dfrac{2\pi}{\lambda}x)[/tex]
[tex]y=A\ sin(2\pi ft+\dfrac{2\pi}{v/f}x)[/tex]
[tex]y=A\ sin\ 2\pi f(t+\dfrac{x}{v})[/tex]
Where
k is the propagation constant
v is the speed of wave
f is the frequency of a wave
[tex]\lambda[/tex] is the wavelength
Hence, this is the required solution.