The speed of a particle executing SHM with amplitude of displacement 5 cm is `3 cm //s ` at a distance 2.5 cm from mean position. What will be its speed at a distance `2.5 sqrt(3)` from mean position?

Explanation

`v= omega sqrt( A^(2) -x^(2))`
At `x= 2.5 cm , v= 3 cm//s `
`implies 3 = omega sqrt( 5^(2)= ( 2.5 )^(2))= ( 5 sqrt(3) omega)/( 2) implies omega = ( 2)/( 5 ) sqrt( 3) rad//s`
Again, `x= 2.5 sqrt( 3) cm , v= ?`
`implies v = omega sqrt( 5^(2) - ( 2.5 sqrt( 3))^(2))`
`= ( 5)/(2) omega = ( 5)/( 2) xx ( 2)/( 5) sqrt( 3) = sqrt(3) cm //s`