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. 2.7 s
C. 2.5 s
D. none of these

Answer

C

Explanation

Let `omega_1` and `omega_2` be the angular frequencis of first and second particle respectively then the phase by which they will proceed in time t is `omega_1t` and `omega_2t`, respectively.
According to the given situation.
`omega_2t-omega_1t=3xx2pi` for `t=45s`
`(2pi)/(T)-(2pi)/(3)=(3xx2pi)/(45)`
`(1)/(T)=(1)/(3)+(1)/(15)=(6)/(15)impliesT=2.5s`