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The time period of oscillation of a particle that executes `SHM` is `1.2s`. The time starting from mean position at which its velocity will be half of its velocity at mean position isA. `2 A`B. `(sqrt(3))/(2)xxA`C. AD. `(A)/(2)` |
Answer» Correct Answer - B `V=(V_(m))/(2)` `omegasqrt(A^(2)-x^(2))=(Aomega)/(2)` `A^(2)-x^(2)=(A^(2))/(4)` `A^(2)-(A^(2))/(4)=x^(2)` `x=sqrt(3)/(2)A`. |
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