In Exercise 9, let us take theposition 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.

In Exercise 9, let us take theposition of mass when the spring is unst

1.	In Exercise 9, let us take theposition 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.
Answer» SOLUTION :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 abovevalues `= 2 sin(2 omega t +0)` ` =2 sin 20 t cm""[therefore omega = 20 rad s^(-1)]`