Is My Solution to the Driven Spring Problem Correct?

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The discussion centers on the correctness of a solution to the Driven Spring Problem. The user defines a Cartesian coordinate system and derives expressions for position and velocity, leading to the formulation of the Lagrangian. The derived Lagrangian is expressed as L = T - V, incorporating kinetic and potential energy terms. A response confirms that the user's approach appears to be correct. The conversation emphasizes the importance of accurate mathematical formulation in solving physics problems.
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Homework Statement
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Relevant Equations
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For this problem,
1718433290547.png

For part(a), I am not sure if I am solving it correctly. I define the usual cartesian x-y coordinate system at the base of the wall. This gives ##x = l_0 + q(t) + x_w(t) = l_0 + q(t) + d\sin(\gamma t)## which implies that ##\dot x = \dot q + d \gamma \cos (\gamma t)##

Therefore ##L = T - V = \frac{1}{2}m(\dot q + d \gamma \cos (\gamma t))^2 - 0.5kq^2##.

Is this please correct?

Thanks!
 
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ChiralSuperfields said:
For part(a), I am not sure if I am solving it correctly. I define the usual cartesian x-y coordinate system at the base of the wall. This gives ##x = l_0 + q(t) + x_w(t) = l_0 + q(t) + d\sin(\gamma t)## which implies that ##\dot x = \dot q + d \gamma \cos (\gamma t)##

Therefore ##L = T - V = \frac{1}{2}m(\dot q + d \gamma \cos (\gamma t))^2 - 0.5kq^2##.

Is this please correct?

Thanks!
Yes, it looks correct.
 
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I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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