In Exercise 9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t=0), the mass is at the maximum streatched position.

Explanation

Amplitude from exercise : `14.9, A = 2 cm`
Angular frequency `omega = sqrt((k)/(m)) = sqrt((1200)/(3)) = sqrt(400) = 20 rad s^(-1)`
Phase in the maximum stretched position `phi =(pi)/(2)rad`
Displacement of block at time t
`x(t) =A sin (omega t+phi)` putting the value ` = 2sin (2omega t+(pi)/(2))= 2 cos 20 t cm`.