Question

A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 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

Answer 1

Answer

given,

Time period= T = 1.5 s

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

v_max=1.00 m/s

we know,

v_ max=A ω  

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt)  =  0.5

$\omega t = sin^{-1}(0.5)$

$\dfrac{2\pi}{T}\times t =0.524$

$\dfrac{2\pi}{1.5}\times t =0.524$

t = 0.125 s

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

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec