A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceslleration at a distance from the mean position 2 cm is `8 cm s^-1`. The maximum speed of the particle is,

Options

A. `8 cm s^-1`
B. ` 12 cm s^-1`
C. ` 16 cms ^-1`
D. ` 24 cms ^-1`

Answer

b

Explanation

Accelaration,`a=8 cm//s^2`
Displacement Amplitude A=6
Since,`a=omega^2y`
`implies 8=omega^2(2)`
`implies omega=2 rads^-1`
`V_max=Aomega`
`=6xx2`
`therefore V_max=12 cm//s`