Question

Two particles A and B execute simple harmonic motions of period T and `5T//4`. They start from mean position. The phase difference between them when the particle A complete an oscillation will be
A. `pi//2`
B. zero
C. `2 pi//5`
D. `pi//4`

Answer 1

Correct Answer - A
Given that, the time period of particle A = T and the time period of particle `B = (5T)/(4)`
Hence, the time difference, `Delta T = (5T)/(4)-T`
`rArr" "Delta T = (T)/(4)`
The relation between phase difference and time difference is
`Delta phi = (2pi)/(T)DeltaT = Delta phi = (2 pi)/(T)xx(T)/(4)rArr Delta phi = (pi)/(2)`