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Let's assume a scenario according to the given picture. If we release the object from point $A$,common sense tells us that it will reach point $B$, then $C$, and then traverse the same path backwards,basically it will do a simple harmonic motion.

But the definition of simple harmonic motion is only proven when restoring force $F=kx$ where $k$ is a constant and $x$ is the distance from equilibrium. Hence the restoring force must vary with the position. Now, according to the figure the force hence acceleration of the object is always pointing towards $B$. So $B$ is our equilibrium or mean position. Now suppose we analyze the position of the object at any random position. And let the angle of the inclined plane with the horizontal in which that position is be $\theta$.

Now as we all know,the only force parallel to the plane is $mg\sin \theta$. So the restoring force towards the equilibrium position or $B$ is $mg\sin \theta$. But all of $m,g,\theta$ are constants here. So the restoring force is a constant regardless of the position and thus it can't be a periodic motion or SHM! Whereas common sense tells us that it is a periodic motion or SHM.

What is the error that we are making in this logic which doesn't match with the common sense scenario?

Please do not mark this question as a homework question. This is a scenario that genuinely came to my mind while thinking about SHMS.

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    $\begingroup$ Your example is not simple harmonic motion $\endgroup$
    – Bob D
    Commented Feb 3, 2023 at 16:04
  • $\begingroup$ The motion may be oscillatory and even periodic, but it cannot be simple harmonic motion $\endgroup$
    – Cross
    Commented Feb 3, 2023 at 16:08

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Your intuition is wrong. You are correct to assume that the motion will be oscillatory, but not all oscillations are harmonic $-$ this is quite a restricted term with a pretty specific meaning, and the situation you depict, as you have shown, does not meet the requirements to be classified as SHM.

If you want a more physical explanation, you can try this: in SHM, the period must be independent of the amplitude. For your example, however, the motion is parabolic on either side of the centre, and this will take longer and longer as the amplitude increases and the object needs to fall down from a higher and higher point.

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