In exercise 14.9 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 (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 u

1.	In exercise 14.9 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 (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)) = 20 rads^(-1)` (a) When mass is at MEAN position Then x =asin `OMEGAT` x = 2 sin 20 t (b) At MAXIMUM stretched position the mass is at extreme right position, HENCE, initial phase is `(pi)/(2)` `therefore x = a sin(omegat + (pi)/(2))` ` = 2 cos omegat = 2 cos 20t` (c ) At maximum compressed position, the mass is at extreme left with initial phase of `(3pi)/(2)` rad. `therefore x = a sin(omegat + (pi)/(2))` ` = 2 cos omegat = 2 cos 20t` (c ) At maximum compressed position, the mass is at extreme left with initial phase of `(3pi)/(2)` rad. `thereforex = a sin (omegat + (3pi)/(2))` `= -a cos omegat = -2cos 20 t` Thus these functions differ in intial phase.