Father pushing child on swing problem

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A father pushes his child on a swing by pulling it back 1 radian and applying a constant force equal to the child's weight until the swing returns to a vertical position. The discussion revolves around deriving the motion equations for the swing, specifically focusing on the approximation that sin(α) is approximately equal to α for small angles. The relevant equation of motion is derived as d²α/dt² + (g/L)α = g, where α is the angle of the swing and L is the length of the swing chain. The user expresses difficulty in solving this second-order differential equation and seeks assistance in finding a solution. The conversation highlights the challenges of applying differential equations to physical problems, particularly after a long hiatus from the subject.
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Homework Statement



A father is about to push his child on the swing. He pulls the swing back by 1 radian, and start pushing the child with a constant force perpendicular to the swing's chain and with a magnitude equal to mg, where m is the mass of the child and the swing seat. The father stops pushing when the swing's chain is again vertical. In the approximation that sinα ≅ α (where α is an angle up to 1 radian), calculate the time the father is pushing. Neglect the weight of swing's chain and the retardation by the air drag.

Homework Equations



Umm, some second order differential equation that I have no clue how to derive?

The Attempt at a Solution



I really have no idea where to start with this one and I only have like 4 hours to get this done :frown: . Any help would be greatly appreciated.
 
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First, please do not use "special symbols". I do not have a Font that allows me to see what you have between sin \alpha and \alpha. I am assuming that you are saying that \alpha is small enough that sin\alpha is approximately equal to \alpha (which is NOT very accurate for \alpha up to 1 radian- sin(1) is only about .84- so I may be wrong about that).

Okay, your basic equation is F= ma, force equals mass times acceleration. At any point, there is a force, -mg, straight down, but the swing chain offsets part of that- the force parallel to motion is -mg sin\alpha so you have m d^2s/dt^2= -mg sin\alpha or approximately m d^2s/dt^2= -mg sin\alpha where s is measured along the arc of the swing. If sin(\alpha) is approximately \alpha, then m d^2s/dt^2= -mg\alpha approximately.
If the swing chain has length L, then s= L\alpha so the equation is d^2\alpha/dt^2= (-g/L) \alpha or md^2\alpha /dt^2+ (mg/L) \alpha= 0.

Now put in the father's push. He is pushing parallel to the arc of motion with magnitude mg so we have md^2\alpha/dt^2+ (mg/L) \alpha= mg. As always with gravity problems, the "m"s cancel and we have d^2\alpha/dt^2+ (g/L) \alpha= g. That's the equation you want to solve.
 
You're a lifesaver, thank you so much for your help!
 
Just when I thought I had this problem, turns out the 2.5 years of no differential equations has caught up to me. Can anyone help me on how to solve d^2\alpha/dt^2+ (g/L) \alpha= g? I've been searching online and through books for the past 2 hours and can't get it :confused:. Thanks.
 
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