Two partical `A and B` execute simple harmonic motion according to the equation `y_(1) = 3 sin omega t` and `y_(2) = 4 sin [omega t + (pi//2)] + 3 sin
Two partical `A and B` execute simple harmonic motion according to the equation `y_(1) = 3 sin omega t` and `y_(2) = 4 sin [omega t + (pi//2)] + 3 sin omega t`. Find the phase difference between them.
A. `(7 pi)/(12)`
B. `(pi)/(12)`
C. `-(pi)/(6)`
D. `(pi)/(6)`
1 Answers
Correct Answer - B
Phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles, i.e. how far one particle leads the other or lags behind the other.
Here, `y_(1)=(1)/(2)sin omega t + (sqrt(3))/(2)cos omega t`
`= "cos"(pi)/(3)sin omega t + "sin"(pi)/(3)cos omega t`
`therefore" "y_(1)=sin(omega t + (pi)/(3))`
and `y_(2) = sqrt(2) sin (omega t + (pi)/(4))`
`therefore` Phase difference, `Delta phi = (pi)/(3)-(pi)/(4)=(pi)/(12)`