Is My Solution to the Driven Spring Problem Correct?

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SUMMARY

The discussion revolves around the Driven Spring Problem, where the user defines the Cartesian x-y coordinate system at the base of the wall. The equations presented include the position equation ##x = l_0 + q(t) + d\sin(\gamma t)## and the derived velocity equation ##\dot x = \dot q + d \gamma \cos (\gamma t)##. The user formulates the Lagrangian as ##L = T - V = \frac{1}{2}m(\dot q + d \gamma \cos (\gamma t))^2 - 0.5kq^2##. The consensus confirms that the user's approach and calculations are correct.

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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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