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A particle executes S.H.M. with amplitude 2 cm. At a distance 1 cm from the mean position magnitudes of the velocity and acceleration are equal. The time period of oscillation is : |
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Answer» `2pisqrt(3)` s `sqrt(r^(2)-y^(2))=omega y` `r^(2)-y^(2)=omega^(2)y^(2)` `IMPLIES""4-1=omega^(2)""implies""omega=sqrt(3)` rad/s. `:.""(2pi)/(T)=sqrt(3)implies""T=(2pi)/(sqrt(3))=(2pi)/(3)sqrt(3)` s. So CORRECT choice is (B). |
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