What is difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is?

Text Solution

`0.5 pi``pi``0.707 pi`Zero

Answer : A

Solution : The displacement equation of particle executing SHM is <br> `x= a cos(omegat+phi)`…(i) <br> Velocity, `v=(dx)/(dt)= -a omega sin(omega+phi)` ….(ii) <br> Acceleration, <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/A2Z_XI_MOC_C02_E01_005_S01.png" width="80%"> <br> `A=(dv)/(dt)= -a omega^(2) cos (omegat+phi)`….(iii) <br> Fig is a plot of Eq.(i) with `phi=0`. Fig.shows Eq.(ii) also with `phi=0`. Fig(iii) is a plot of Eq.(iii). It should be noted that in the figures the curve of `v` is shifted (to the left) from the curve of x by one-quarter period `(1/4T)`. Similarly, the acceleration curve of A is shifted (to the left) by `1/4T` relative to thevelocity curve `v`. This implies that Velocity is `90^(@)(0.5pi)` out phase with the displacement and the acceleration is `90^(@)(0.5pi)` out phase with the velocity but `180^(@)(pi)` out of phase with displacement.