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 compressed position. In what way to these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

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 compressed position. In what way to these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
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 inthe maximum compressed position `phi = (3pi)/(2) rad` Displacement of block at time t `x(t) = A SIN (omega t+phi)` ` = 2sin (2omega t+(3pi)/(2))= -2cos 20 t cm`. Hence, in all these three CASES, amplitude and frequency are same but differ only in initial phase because motions starts from different positions.