A particle moving in simple harmonic motion with a period T

A particle moving in simple harmonic motion with a period T 1.5 S passes through the equilibrium point at time r0 * 0 with a velocity of 1.00 m/s to the right. A time t later, the particle is observed to move to the left with a velocity of 0.50 m/s. (Note the change in direction of the velocity.) the smallest possible value of the time t is

Solution

If it\'s moving through equilibrium point at t₀=0 with v=1.0m/s

then vmax=1.00m/s vmax=A

So, -0.50= -1.00sin (t)

sin-1(0.50)= 2pi/T x t

solving this we get t= 0.125 sec.-----------------1

we have time period T=1.5 it is the time to complete one oscillation(means from eq to right,then left,then eq,then left,then from right to eq.)

so time from eq to left= 1.5/4=0.375sce----------------------2

the smallest value of time is given by adding 1 and 2

smallest value of time=0.375+0.125=0.50sec

Reference https://www.physicsforums.com/threads/particle-moving-in-simple-harmonic-motion.624702/