Two waves are represented by the equations
`y_(1)=a sin (omega t+kx+0.785)`
and `y_(2)=a cos (omega t+kx)`
where, x is in meter and t in second
The phase difference between them and resultant amplitude due to their superposition are

Options

A. `45^(@)` and 1.84 a
B. `30^(@)` and a
C. `30^(@)` and 2a
D. `45^(@)` and a

Answer

A

Explanation

`y_(1)= a sin (omega t+kx+0.57)`
`:.` Phase `phi_(1)=omegat+kx+0.57`
`y_(2)=a cos (omega t+kx)`
`=a sin (omega t+kx +pi/2)`
`:.` Phase, `phi_(2)=omegat+kx+pi/2`
Phase diferent, `Delta phi=phi_(2)-phi_(1)`
`=[omega t+kx+pi/2]-(omegat+kx+0.57)`
`=pi/2 -0.785`
`=(1.57-0.785) rad =0.785` rad
`=pi/4 rad = 45^(@)`
Resultant amplitude, `a=2 a "cos" phi/2=2a " cos"pi/8=1.84 a`