Asked by student1618
Consider an ideal spring that has an unstretched length ℓ and a spring constant k. Suppose the spring is attached to a mass m that lies on a horizontal frictionless surface. The spring-mass system is compressed a distance of x_0 from equilibrium and then released with an initial speed v_0 toward the equilibrium position.
What is the maximum distance the spring will be stretched?
I tried conservation of energy but v_0 can't be used in the solution so I'm at a lost as to what to do.
What is the maximum distance the spring will be stretched?
I tried conservation of energy but v_0 can't be used in the solution so I'm at a lost as to what to do.
Answers
Answered by
Damon
x = a sin w t + b cos wt
v = a w cos wt - b w sin w t
omega = w = sqrt(k/m)
at t = 0
x = -xo
v = dx/dt = vo
- xo = b
+ vo = a w or a = vo/w
so x = (vo/w) sin w t - xo cos wt
now I could say
magnitude (or amplitude) = sqrt (vo^2/w^2 + xo^2) which is max motion from equilibrium
= sqrt (vo^2 m/k + xo^2)
v = a w cos wt - b w sin w t
omega = w = sqrt(k/m)
at t = 0
x = -xo
v = dx/dt = vo
- xo = b
+ vo = a w or a = vo/w
so x = (vo/w) sin w t - xo cos wt
now I could say
magnitude (or amplitude) = sqrt (vo^2/w^2 + xo^2) which is max motion from equilibrium
= sqrt (vo^2 m/k + xo^2)
Answered by
student1618
but how to I eliminate v_0?
Answered by
student1618
never mind I typed it wrong. Thanks!
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