Two particles are executing simple harmonic of the same amplitude (A) and frequency `omega` along the x-axis . Their mean position is separated by distance `X_(0)(X_(0)gtA). If the maximum separation between them is (X_(0)+A), the phase difference between their motion is:

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Answer

A

Explanation

Two particle ar executing S.H.M.

Distance between mean position ` = X_(0)(X_(0) gt A)`
Maximum separation between them is `(X_(0) + A)`
Let position of oscillation from reference CD line
`X_(1) = A sin (omegat)`
`X_(2) = X_(0) + A sin(omegat + phi)`
`X_(2) = X_(1) = X_(0) + A[sin (omegat + phi) - sin omegat]`
` =X_(0) - A [2cos((2omegat + phi)/(2)) sin ((theta)/(2))-]`
`X_(2) - X_(1) = X_(0) - 2 A sin ((theta)/(2)) cos [omegat + ((phi)/(2))]`
`x_(max) = X_(0) + 2 A sin ((phi)/(2)) cos [omegat + ((phi)/(2))]`
`X_(max) = X_(0) + 2 A sin ((phi)/(2)) {cos (omegat + (phi)/(2)) = -1}`
`X_(0) + A = X_(0) + 2 A sin ((phi)/(2))`
`2A sin ((phi)/(2)) = A`
`sin ((theta)/(2)) = (1)/(2)`
`phi = (pi)/(3)`