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.

Options

A. `(pi)/(2)`
B. `tan^(-1)((4)/(5))`
C. `tan^(-1)((3)/4))`
D. None of these

Answer

B

Explanation

Representing a quantity with phase `(omegat)` along `X`-axis any quantity with a phase `(omegat + phi)` can be represented by a vector making an angle `phi` with a phase `(omegat - phi)` can be represented by a vector making an angle `phi` with x- axis in the clokwise sense

Phase difference `phi = tan ^(-1) ((4)/(5))`
Altmetively given `y_(1) = 3 sin omega t `....(1)
And `y_(2) = 4 sin (omega t + (pi)/(2)) + 3 sin omega t`
`= 4 cos omega t + 3 sin omega t = 5 [(4)/(5) cos omega t + (3)/(5) sin omega t]`
`= 5[sin phi cos omega t + cos omega t sin omega t]`
`= 5 sin (omega t + phi)`

`cos phi = (3)/(5) sin phi = (4)/(5) and tan phi = (4)/(5)`
From (i) and (ii) the result follows