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 maximum streatched 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 maximum streatched 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)` Phase in the maximum stretched position `phi =(pi)/(2)rad` Displacement of block at time t `x(t) =A sin (omega t+phi)` putting thevalue ` = 2SIN (2omega t+(pi)/(2))= 2 cos 20 t cm`.