A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position the velocity of the particle is 10 cm/ s . The distance of the particle from the mean position when its speed becomes 5 cm/s is

Options

A. `2(sqrt(3)) cm`
B. `2(sqrt(5))cm`
C. `sqrt(5) cm`
D. `sqrt(3) cm`

Answer

A

Explanation

At the mean position, `v=A omega`
`therefore 10=4xx omega therefore omega=(5)/(2) rad//s`
and `because v=omegasqrt(A^(2)-x^(2))`
`therefore v^(2)=omega^(2)(A^(2)-x^(2)) therefore 25=(25)/(4)(16-x^(2))`
`therefore 4=16-x^(2) therefore x^(2)=16-4=12 therefore x=2 sqrt(3) cm`