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:

Options

A. 3A
B. `Asqrt3`
C. `(7A)/(3)`
D. `(A)/(3)sqrt41`

Answer

C

Explanation

`c v = omega(sqrt(A^(2) - y^(2)))`
At, `y = (2A)/(3) " " v = omega[A^(2) - (4A^(2))/(9)]^(1//2)`
`v = omega[(5A^(2))/(9)]^(1//2) rArr v^(2) = omega^(2)((5A^(2))/(9))" "…(i)`
Let new amplitude be A. and final velocity be 3v
`3v = omega sqrt(A^(2) - ((2A)/(3))^(2))`
Squaring on both sides, and from equation (i)
`9 (omega^(2)(5A^(2))/(9)) = omega^(2)(A^(2) - (4A^(2))/(9))`
`5A^(2) = A^(2) - (4A^(2))/(9)`
`A^(2) = (49)/(9)A^(2)`
`A. = (7)/(3)A`