InterviewSolution
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InterviewSolution
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A particle performing SHM has time period ` (2pi)/(sqrt(3))` and path lenght 4 cm. The displacement from mean position at which acceleration is equal to velocity isA. 0 cmB. 0.5 cmC. 1 cmD. 1.5 cm |
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Answer» Correct Answer - C Velocity ` v=omegasqrt(A^(2)-x^(2))` and acceleration ` =omega^(2)x` Given, `omega sqrt(A^(2)-x^(2))=omega^(2)x` or ` sqrt(A^(2)-x^(2))=omega^(2)x` ......(i) Given , ` T=(2pi)/(sqrt(3))` and ` omega=(2pi)/(T)=sqrt(3)` Substituting the value of ` omega ` in Eq(i) , we get ` sqrt(A^(2)-x^(2))=sqrt(3)x` `implies A =2 x ` As amplitude`=("path length")/(2)=2 cm ` `implies x=1 cm ` |
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