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Two particles A and B execute simple har...

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

`0`

C

`2pi//5`

D

`pi//4`

Text Solution

Verified by Experts

The correct Answer is:
C
Phase difference `(Deltaphi)=(2pi)/(T)xx`time difference
`Deltaphi_(1)=(2pi)/(T)xxT=2pi`
`Deltaphi_(2)=(2pi)/((5T)/(4))*T=(8pi)/(5)`
`therefore Delta phi=Deltaphi_(1)-Deltaphi_(2)=2pi-(8pi)/(5) implies Deltaphi=(2pi)/(5)`.
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