A mass 2kg is attached to the spring of spring constant `50 Nm^(-1)`. The block is pulled to a distance of 5 cm from its equilibrium position at x= 0 on a horizontal frictionless surface from rest at t=0. Write the expression for its displacement at anytime t.

Explanation

Spring block system is shwon as in figure and it executes SHM with amplitude of 5 cm from mean position.

Here, spring constant `k= 50 N"/"m`, amplitude A= 5cm, mass attached m = 2 kg.
Angular frequency `omega = sqrt((k)/(m))= sqrt((50)/(2))= 5 rad"/"s`.
Displacement of block at time t,
`y(t) = A sin (omega t+ phi)," where "phi=` initial phase at time t=0
`y(0)= A sin (phi)`
`A= A sin phi" "[therefore " at time "t=0, y= +A]`
`therefore 1= sin phi`
`therefore phi = (pi)/(2) rad`
Required equation, `y(t) = A sin(omega t+phi)`
`= 5sin (omega t+(pi)/(2))`
`therefore y(t) = 5 cos omega t`
`y(t)= 5 cos 5t` is a required equation where t is in second and y is in cm.