Two SHMs are respectively represented by `y_(1)=a sin (omegat-kx) and y_(2)=b cos(omegat-kx)`. The phase difference between the two is

Options

A. `45^(@)`
B. `90^(@)`
C. `60^(@)`
D. `30^(@)`

Answer

B

Explanation

`y_(1) = a sin (omega t - kx)`,
`y_(2) = b cos (omega t-(k)/(x))=b sin (omega t-(k)/(x)+pi//2)`
`therefore` Phase difference `= (omega t-(k)/(x)+pi//2)-(omega t - kx)`
`= pi//2 = 90^(@)`