sin ω t + B cos ωt dx dt = A ω cos ωt − B ω sin ωt d x dt = − ω (A sin ωt+ B cos ωt) d x dt = − ω x This differential equation is similar to the differential equation of SHM (equation . ). Therefore, x = A sin ωt + B cos ωt represents SHM. (ii) x =A sin ωt + B cos2 ωt dx dt = A ω cos ωt − B ( ω ) sin2 ωt d x dt = − ω (A sin ωt+ B cos ωt) d x dt ≠ − ω x This differential equation is not like the differential equation of a SHM (equation .
). Therefore, x = A sin ωt + B cos ωt does not represent SHM. (iii) x = A e iωt dx dt = A i ωe iωt d x dt =− A ω e iωt =−ω x ( ∴ i =– ) This differential equation is like the differential equation of SHM (equation . ).
Therefore, x = A e iωt represents SHM. (iv) x = A ln ω t dx dt = A t t ω ω d x dt = - A t ⇒ d x dt ≠−ω x This differential equation is not like the differential equation of a SHM (equation . ). Therefore, x = A ln ω t does not represent SHM.
EXAMPLE . Consider a particle undergoing simple harmonic motion. The velocity of the particle at position x is v and velocity of the particle at position x is v . Show that the ratio of time period and amplitude is x x v x v x π o o x t t o a acceleration velocity Displacement x = A sin ω t v = ω A cos ω t a = – ω A sin ω t t Figure .
Variation of displacement, velocity and acceleration at different instant