# Simple Harmonic Motion

The red particle is going round the circular path in anti-clock wise sense (considered positive) with a uniform speed and its x-projection and y - projection are shown in cyan and orange respectively. If the angular position of the red particle is denoted by θ then its x and y co-ordinates are R cos θ and R sin θ (With x to the right and y upward and origin at the center of the circle). But since the motion of the particle round the circle is uniform θ   = ω   t  + δ , where ω is the angular speed and δ is the initial angular position of the red particle. In the applet δ is set to zero. θ is shown by the small blue filled arc at the center. In the applet the phase is reset to zero after every revolution. In practice it would increase continuously, going up by 2π radians for every revolution. The cyan particles motion can be represented as x   =   A   Cos   (ω   t   +   δ) and that of orange particle as y   =   A   Sin(ω   t   +   δ) where A is the amplitude of SHM which is equal to the radius of the circle and ω t+δ is called the phase. δ is the initial phase. You should be able to see that eliminating time from x and y equations gives you the equation of the circle. Please note that even if θ varies in a more complicated manner this would be true. But the x and y component motions then would not be simple harmonic. Do notice how the velocity vectors of the cyan and orange particles vary. You should be able to see that the velocity vectors diminish more rapidly when the particles are farther from the center, indicating greater acceleration when farther from the center of the circle. Recall that in SHM acceleration is proportional to the negative of displacement. That negative sign comes in because when displacement is positive and increasing, the velocity vector would be positive and decreasing. and if displacement is positive and decreasing, the velocity vector would be negative and increasing. You can check the cases when displacement is negative.

Surendranath .B.