Spring Constant > Create second order equation
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 secondorder equation you wanted. To solve it, you need a pair of initial conditions, which the problem provides you: $x(0) = 0$ (since the initial position is $0$) 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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Mhsmith21 over 1 year
The Problem:
A frictionless spring with a 10kg 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)

Mhsmith21 almost 10 yearsI think I did it, can you verify my setup please

user7530 almost 10 yearsLooks good to me!