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Homework Statement
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
Homework Equations
V= -Aω sin(ωt)
The Attempt at a Solution
If it's moving through eq point at t0=0 with v=1.0m/s then vmax=1.00m/s
vmax=Aω So,
-0.50= -1.00sin (ωt)
sin-1(0.50)= 2pi/T x t
pi/6=2pi/T x t
t= pi/6 x T/2pi = 0.125 sec.
If T=1.5 seconds and it is moving to the right initially then half way through the motion is t=.75 so does that mean that it only begins to move left at t=0.75 seconds after it has reached it max point to the right? Therefore the 0.125 second value I got from the above equation should be added to 0.75 seconds to get the minimum value for t when the particle is moving left at 0.50 m/s?
I feel like I'm missing something. There were no examples of this type of problem done in discussion, not much explanation in lecture, and the book is useless so I feel like I'm going at this pretty blind. Any help explaining this problem to me would be great! Thanks!