Asked by Kate
A mass on the end of a spring is stretched a distance Xo from equilibrium and released. At what distance form equilbrium will it have (a) velocity equal to half its maximum velocity and (b) acceleratipon equal to half its maximum acceleration ?
I understand (b) that's just have Xmax but I do not understand how to go about (a)
the back of the book tells me .866 Xmax which I don't see were it came from
I understand (b) that's just have Xmax but I do not understand how to go about (a)
the back of the book tells me .866 Xmax which I don't see were it came from
Answers
Answered by
Damon
x = Xo cos wt where w = 2 pi f
v = -Xo w sin w t
a = -Xo w^2 cos wt = -w^2 x
Max v = -Xo w
1/2 max v = -.5 Xo w
at what x?
-Xo w sin wt = -.5 Xo w
when sin wt = .5
that is when wt = 30 degrees
so what is cos wt then?
cos 30 degrees = .866
so
x = Xo cos 60 = .866 cos 60
v = -Xo w sin w t
a = -Xo w^2 cos wt = -w^2 x
Max v = -Xo w
1/2 max v = -.5 Xo w
at what x?
-Xo w sin wt = -.5 Xo w
when sin wt = .5
that is when wt = 30 degrees
so what is cos wt then?
cos 30 degrees = .866
so
x = Xo cos 60 = .866 cos 60
Answered by
Damon
I mean .866 Xo
in the last line
in the last line
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