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The Problem:

A frictionless spring with a 10-kg mass can be held stretched 1 meters beyond its natural length by a force of 40 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2.5 m/sec, find the position of the mass after t seconds.

How do I convert this to a second order equation and then solve it?

Spring constant would be 40nm = k * 1 meter -> k = 40

Thanks

ATTEMPTS BASED ON COMMENTS/ANSWERS

10x'' + 40x = 0, when x(0) = 1 and x'(0) = 2.5

10r^2 + 40 = 0

10r^2 = -40

r^2 = -4

r = +- 2i

y  = Acos(2x) + Bsin(2x) = homogenous solution

y' = -2Asin(2x) + 2Bcos(2x)

1 = Acos(2*0) + Bsin(2*0) = A

2.5 = -2Asin(2*0) + 2bcos(2*0) = 2B


y = cos(2x) + 5/4sin(2x)
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Hooke's law tells you that $$F = -k x$$ and Newton's second law, that $$F = ma = m\ddot{x}.$$

Putting these pieces together you get that $$m\ddot{x} = -kx$$ or $$10\ddot{x} = -40x,$$ which is the second-order equation you wanted. To solve it, you need a pair of initial conditions, which the problem provides you: $x(0) = 1$ and $\dot{x}(0) = 2.5$ (I'm assuming the push gives it a velocity in the direction of increasing distention; the problem really should be more precise.) If you need help solving the ODE, you can ask, but first show the work of your attempt.

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  • $\begingroup$ I think I did it, can you verify my setup please $\endgroup$ – Mhsmith21 Dec 12 '13 at 2:27
  • $\begingroup$ Looks good to me! $\endgroup$ – user7530 Dec 12 '13 at 2:29

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