Correct Answer - D
The given transverse harmonic wave equation is
`y=3sin(36t+0.018x+pi/4)`..(i)
As there is positive sign between t and x terms therefore the given wave is travelling in the negative x direction. The standard transverse harmonic wave equation is
`y=asin(omegat+kx+phi)`..(ii)
Comparing (i) and (ii), we get
`a=3 cm,omega=36rads^(-1),k=0.018radcm^(-1)`
`therefore` Amplitude of wave, a = 3 cm
Frequency of the wave, `upsilon=omega/(2pi)=36/(2pi)=18/piHz`
`v=omega/k=(36rads^(-1))/(0.018cm^(-1))=2000cms^(-1)=20ms^(-1)`