In Exercise 14.9, let us take the position of mass when the spring is unstreched 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 (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?

In Exercise 14.9, let us take the position of mass when the spring is

1.	In Exercise 14.9, let us take the position of mass when the spring is unstreched 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 (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?
Answer» Solution :`a = 2 cm, omega = SQRT((k)/(m)) = sqrt((1200)/(3)) s^(-1) = 20s^(-1)` (a) SINCE time is measured from mean position, `x = a sin omega t = 2 sin 20t` (b) At the maximum stretched position, the body is at the extreme RIGHT position. The initial phase is `(pi)/(2)` `:. x = a sin (omega t +(pi)/(2)) = a cos omega t = 2 cos 20 t` (c) At the maximum compressed position, the body is at the extreme left position. THEINITIAL phase is `(3pi)/(2)` `:. x = a sin (omegat +(3pi)/(2)) =- a cos omega t =- 2 cos 20 t` Note: The functions neither DIFFER in amplitude nor in frequency. They differ only in initial phase.