Obtain instantaneous velocity of a particle executing SHM.

Explanation

Intantaneous velocity of SHM particle is the rate of change of displacement for a time interval.
Suppose the displacement of SHM particle at time t with amplitude A and angular speed `omega` is
`x(t)= A cos (omega t+ phi)" ""……"(1)`
where `phi` is the initial phase
Differentiating equation (1) respect to t gives the instantaneous velocity
`therefore v(t)= (d[x(t)])/(dt) = (d)/(dt)[A cos (omega t+ phi)]`
`therefore v(t)= -A omega sin (omega t+ phi)`
but `sin(omega t +phi)= pm sqrt(1-cos^(2) (omega t+ phi))`
`v(t)= (-A omega) (pm sqrt(1-cos^(2) (omega t+ phi)))`
`= pm omega sqrt(A^(2)-A^(2) cos^(2)(omega t+ phi))`
`= pm omega sqrt(A^(2)-x^(2)(t))`
Generally `v= pm omega sqrt(A^(2) - x^(2))`
Special Cases :
(1) At mean position, x=0
Velocity `v= pm omega A`
In +x-direction, v is positive and in -x-direction, v is negative.
Hence, maximum velocity of SHM particle `v_("max")= A omega`
(2) At extreme point, for velocity `x= |A|`,
`v= pm omega sqrt(A^(2)-A^(2))" "[therefore |A|^(2)= A^(2)]`
`therefore v= 0`
The velocity of SHM particle at extreme points is zero.