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A particle performing simple harmonic mo...

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

A

`2.8s`

B

`2.7s`

C

`2.5s`

D

`3.2s`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta phi=((2pi)/(T_(2))-(2pi)/(T_(1)))DeltaT`
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