A particle performing simple harmonic motion having time period 3 s is in phase with another particle which also undergoes simple harmonic motion at `t=0`. The time period of second particle is T (less that 3s). If they are again in the same phase for the third time after 45 s, then the value of T will be

Options

A. 2.8 s
B. 2.7s
C. 2.5s
D. 3.2s

Answer

C

Explanation

`Delta phi = ((2pi)/(T_(2)) - (2pi)/(T_(1)))Delta t`
After `Delta t = 45s`, they are again in phase for the 3rd time
`Delta phi = 3 xx 2pi = 6pi`
`6pi = ((2pi)/(T) - (2pi)/(3))(45) rArr T = 2.5`