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 mean 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)`
Initial phase at mean position `phi = 0`
Displacement of block at time t
`x(t) = A sin(omega t+phi)` putting the above values
`= 2 sin(2 omega t +0)`
` =2 sin 20 t cm" "[therefore omega = 20 rad s^(-1)]`