a constant. Therefore, dy dt = A ω cos ω t ( . ) Using trigonometry identity, sin ωt + cos ωt = ⇒ cos ωt = - sin ω t we get t ω ω sin From equation ( . ), sin ω t y y − ω y ω ( .
) From equation ( . ), when the displacement y = , the velocity v = ωA (maximum) and for the maximum displacement y = A, the velocity v = (minimum). As displacement increases from zero to maximum, the velocity decreases from maximum to zero. This is repeated.
Since velocity is a vector quantity, equation ( . ) can also be deduced by resolving in to components. Acceleration The rate of change of velocity is acceleration. Taking derivative of equation .
with respect to time, a dv dt d dt A t ( ) ω ω cos a t y =− =− ω ω ω sin ( . ) ∴ a d y dt y =− ω ( . ) From the Table . and figure .
, we observe that at the mean position Table . Displacement, velocity and acceleration at different instant of time. Time ω t π π π π Displacement y = A sin ω t − A Velocity v = A ω cos ω t A ω − A ω A ω Acceleration a =- A ω sin ω t − A ω A ω - - - - Unit Oscillations ( y = ), velocity of the particle is maximum but the acceleration of the particle is zero. At the extreme position ( y = ± A ), the velocity of the particle is zero but the acceleration is maximum Aω acting in the opposite direction.
EXAMPLE . Which of the following represent simple harmonic motion? (i) x = A sin ω t + B cos ωt (ii) x = A sin ωt+ B cos ωt (iii) x = A e iωt (iv) x = A ln ωt Solution (i) x = A