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It all started with this question.

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There are three different ways to solve this but one way is using kepler's second law. $\frac{dA}{dt}=\frac{L}{2m}.$ This applies because angular momentum is conserved about the origin since net net torque about that point is zero. So this turns into basic integration problem. $\int_0^AdA\,= \frac{L}{2m}\int_0^T\,dt$. Where $T$ is the time period. Time period happens to be $2\pi \sqrt{\frac{m}{k}}$. And writing $L$ as $m\frac{J}{m}A$ we get option D to be right. The only part that bugs me is that why is time period still the same. If you look at the derivation for time period we solve that SHM differential equation, now if you add this motion in $y$-axis doesn't this mess everything up. What is the derivation vectorially or whatever for time period to be same. Is there some general prinicple I am missing?

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  • $\begingroup$ If you show all three ways to solve the problem, I am sure that in the process, you will see a general principle that connects all three ways of looking at the problem. $\endgroup$
    – KDP
    Commented Feb 17 at 13:09

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Circular motion and SHM are very closely related. If you suspend a weight from a string and set the weight in circular motion as looked from above, it will look like linear SHM from the side and will of course have the same time period. Linear SHM is simply circular motion looked at from another angle. This animation of Scotch Yoke from Wikipedia also demonstrates the idea:

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The slotted rectangular block in the middle is also undergoing linear SHM when looked at from the side.

You can also see the equivalence from the phase space diagram (from the same Wikipedia article above) of the SHM, which is a plot of displacement from the centre versus velocity. The result is a circle. In both cases the motion is due to the same constant force acting towards the centre of the motion.

Now the period of circular motion is given by $T = 2 \pi R /v = 2\pi /\omega$ where $\omega = d \theta/dt$ is obtained from the phase space diagram. Now it is just a case of replacing $\omega$ with $\sqrt{k/m}$ to get $T = 2 \pi / \omega = 2 \pi / \sqrt{k/m}$.

It's worth noting that although we have been talking about circular motion here, the path of the particle connected to the spring will in fact be an ellipse, but Kepler's second law still works the same for an ellipse so that is not a problem.

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I am pretty sure you can treat this as a superposition of 2 SHMs one in the x direction and one in the y direction. You can derive the equation of ellipse from this too. Calculating time period is trivial.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
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