Two transverse waves A and B superimposed to produce a node at x = 0. If the equation of wave A si `y= a cos (kx-omegat)`, then the equation of wave B is

Options

A. `y=asin(omegat+kx)`
B. `y=asin(omega-kx)`
C. `y-asin(omegat+kx)`
D. `y=-asin(omegat-kx)`

Answer

C

Explanation

The equation of a wave is
`Y = a sin (omega t-kx)` …(i)
Let the eqyations of another wave be
Either `Y = a sin (omega t-kx)` …(ii)
or , `Y = - a sin (omega t-kx)` …(iii)
If (i) superposes on with (ii), then we get :
`Y = 2a coskx sin omega t` ...(iv)
If (i) superposes on with (iii) , then we get : `Y = -2a sin kx cos omega t` ..(v)
After putting x = 0 in (iv) and (v) , respectively , we get :
y=0 at x=0 For equation (v)
Hence , (iii) is an equation of the unknown wave .