1.

A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is A and its time period is T. At any instant, its speed is half that of the maximum speed. What is the displacement of the particle at the that point?

Answer»

`(2A)/sqrt(3)`
`(3A)/sqrt(2)`
`(sqrt(2)A)/3`
`(sqrt(3)A)/2`

Solution :Maximum velocity `v_("max") = AOMEGA`
ACCORDING to QUESTION, `v_("max")/2 = (Aomega)/2 = omega sqrt(A^(2) -y^(2))`
`A^(2)/4 = A^(2) -y^(2) rArr y^(2) = A^(2) -A^(2)/4 rArr y=(sqrt(3)A)/2`


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