1.

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance `(2A)/3` from equilibrium position. The new amplitude of the motion is:A. `(A)/(3)sqrt(41)`B. `3A`C. `Asqrt(3)`D. `(7A)/(3)`

Answer» Correct Answer - D
Applying conservation of energy
`(1)/(2) m omega^(2) ((2A)/(3))^(2) + 9 xx (1)/(2) m omega^(2) [A^(2) -((2A)/(3))^(2)]`
`= (1)/(2) m omega^(2) A_(1)^(2)`
`rArr A_(1) = (7A)/(3)`


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