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What is the definition of simple harmonic motion? The motion of a particle where its displacement is given by a sinusoidal function, that is: $$x=A\sin(ωt+φ)$$ or the total force that is acting on the particle is: $$ΣF=-kx$$

Also why we need to show that for a particle undergoing simple harmonic motion the force is in the form of the second equation? At some instant some force (other than the initial forces) may act on the system so the 2nd equation isn't satisfied. Would it better to say that instantaneously the particle undergoes simple harmonic motion? In other words can we say that for a net force that is time dependent (more general case) $ΣF(t)=ma$ at some instant $ΣF(t_{shm})=-kx$ the particle will undergo simple harmonic motion?

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If you solve the equation of motion for the system (Newton's second law) with a time-independent force that is proportional to the displacement, $F=-kx$, then the solution is a sinusoidal function. In this sense, the two initial statements that you have made are equivalent. Yet another equivalent way of defining harmonic motion is to describe it in terms of the potential energy, which is proportional to the displacement squared: $$ V(x)=\frac{1}{2}kx^2. $$ You can check that from this potential you obtain the force you wrote by calculating the force as the negative of the gradient of the potential.

For your second question, if at some point in time you do change the forces acting in your system, the equation of motion will change accordingly, and the resulting motion of the particle will no longer be harmonic. So yes, if you insist on making this distinction, you can say that your system is, at present, undergoing simple harmonic motion, and that in the future it may change. But this is true for any type of motion, not only simple harmonic motion. If you change the force, then the motion will change.

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  • $\begingroup$ Thanks for the answer. What if we had $F=-k\vec{r}$. Would these be simple harmonic? $\endgroup$ – Antonios Sarikas Jun 7 '20 at 16:57
  • $\begingroup$ You can also have harmonic motion in 3D. If the oscillator is isotropic, then using spherical coordinates the problem simplifies to a 1D problem. More interesting is the case when the oscillator is not isotropic. $\endgroup$ – ProfM Jun 7 '20 at 17:31

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