InterviewSolution
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InterviewSolution
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A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds isA. `(1)/(2pisqrt(3))`B. `2pisqrt(3)`C. `(2pi)/(sqrt(3))`D. `(sqrt(3))/(2pi)` |
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Answer» Correct Answer - C (c) Velocity , `v=omegasqrt(A^(2)-x^(2))` and acceleration `=omega^(2)x` Now given, `omega^(2)x=omegasqrt(A^(2)-x^(2))` `impliesomega^(2).1=omegasqrt(2^(2)-1^(2))` `implies omega=sqrt(3)` `therefore T=(2pi)/(omega)=(2pi)/(sqrt(3))` |
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