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Two particle are executing simple harmon...

Two particle are executing simple harmonic motion. At an instant of time t their displacement are `y_(1)=a "cos"(omegat)` and `y_(2)=a "sin" (omegat)`
Then the phase difference between `y_(1)` and `y_(2)` is

A

`120^(@)`

B

`90^(@)`

C

`180^(@)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `y_(1)=a "cos "omegat=a "sin "(omegat+(pi)/(2))`
`y_(2)=a " sin "omegat`
`therefore Deltaphi=omegat+(pi)/(2)-omegat=(pi)/(2)=90^(@)`
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