Asked by Dib
What is the velocity of the simple harmonic motion problem?
Position of the graph is
y= .4*cos(2pi*1/3*x)+1.2
I need to find the velocity as well as the acceleration. I have worked through it and keep getting that the velocity is 0. I don't believe that is correct.
Position of the graph is
y= .4*cos(2pi*1/3*x)+1.2
I need to find the velocity as well as the acceleration. I have worked through it and keep getting that the velocity is 0. I don't believe that is correct.
Answers
Answered by
Damon
You do not have time in your equation, only a solid unmoving cosine function.
Do you mean t instead of x ?
y = .4 cos[ (2 pi/3) t ] + 1.2 ?
if so then
dy/dt = v = -.4(2 pi/3) sin[ (2 pi/3) t ]
a = d^2y/dt^2 = -.4(2 pi/3)^2 cos[ (2 pi/3) t ]
which is
a = - (2pi/3)^2 * y
Do you mean t instead of x ?
y = .4 cos[ (2 pi/3) t ] + 1.2 ?
if so then
dy/dt = v = -.4(2 pi/3) sin[ (2 pi/3) t ]
a = d^2y/dt^2 = -.4(2 pi/3)^2 cos[ (2 pi/3) t ]
which is
a = - (2pi/3)^2 * y
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