Show that the motion of a particle represented by y= sin omega t- cos omega t is simple harmonic with a period of (2pi)/(omega).

Show that the motion of a particle represented by y= sin omega t- cos

1.	Show that the motion of a particle represented by y= sin omega t- cos omega t is simple harmonic with a period of (2pi)/(omega).
Answer» SOLUTION :Displacement `y= sin omega t - cos omega t` `therefore y= sqrt(2) [(1)/(2) sin omega t- (1)/(sqrt(2)) cos omega t]` `therefore y= sqrt(2) [cos (pi)/(4)sin omega t - sin (pi)/(4) cos omega t]` `therefore y= sqrt(2) sin (omega t-(pi)/(4))` comparing this euqation with general EQUATION of SHM `y= A sin (omega t + phi)`, Amplitude A= 2 unit ANGULAR frequency, `omega = (2pi)/(T)` `therefore T= (2pi)/(omega)` `therefore omega = (2pi)/(T)` `therefore T= (2pi)/(omega)` Given function represent SHM with a periodic time `T= (2pi)/(omega)`.