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I was looking at different kinds of SHMs on $x$-axis and I was wondering if the situations of positions of the particle at different times are similar for different SHMs as the most general SHM (the one in which particle starts from the origin at $t=0 $) I have attached an image to make this more clear (it's just to describe what I mean by 'different SHMs'.

NOTE: these obviously aren't 'different' SHMs the only difference is when and where they start. But the mean position remains the Origin.

FIGURE


As we know for the SHM in which particle starts from the origin at t=0 towards positive x, it's positions are quite symmetrically determined at different times. That is, at $t=\frac{T}{4}$ , the particle will be at $+A$ where A is the amplitude and T is the time period.
At $t=\frac{T}{2}$, the particle will again be at the origin moving towards negative x.
At $t=\frac{3T}{4}$ the particle will be at $-A$
and finally at $t=T$ the particle will again be at origin completing 1 cycle.


Now my question is will this be the same for the 'other SHMs'? That is, if we consider the first image,
will this particle's displacement be $-A/2$ at $t=T/4$ ? and would it again follow some symmetrical displacement at these specific times?
Would the time period of this even be T?


NOTE: In all the cases we are taking here, the mean position is origin.
I know I might be missing out on a lot of basic points but I'm really confused and need some clear explanation about this. Please explain this on the basis of the first image only.

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I think that the answer to your question is: no.

You seem to be suggesting that the body might cover equal distances in equal time intervals from whatever point in its cycle we start our time interval (or at least that it covers distances of $A$ if we start a time interval of $T/4$ not just at $x=A,\ x=0,\ \text{or}\ x=-A$ but also at certain other points).

Everything is contained in the equation $$x= A\ \cos \left(\frac{2\pi}{T}t + \epsilon\right).$$ If we agree to call $t=0$ when the body has its maximum $x$-wise displacement, $A$, we can put $\epsilon =0$.

For $x=\frac A2$ we find that the smallest positive value of $t$ is $t=\frac16 T$.

For $x=-\frac A2$ we find that the smallest positive value of $t$ is $t=\frac13 T$.

So the time interval for the body to go from $x=\frac A2$ to $x=-\frac A2$ is $\frac13 T-\frac16 T=\frac16T.$

The reason why this is less than $\frac 14 T$ is that, for a body doing shm, the nearer the body is to its equilibrium position, the faster it moves. The relationship between $x$ and $t$ is non-linear.

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  • $\begingroup$ Alright... so we can see that kind of symmetry only in the initial case i was talking about right? like... I'm not saying that the particle covers equal distances in specific intervals. I'm just saying that we observe that after $\frac{T}{4}$ we see it at +A and so on... But we cant apply this to other SHMs starting from some other position right? That is, the displacements won't be specifically defined. $\endgroup$ Commented Mar 11, 2021 at 15:13
  • $\begingroup$ Also... what about the time period? Would the time period of the First picture also be T? That is, where would we observe that particle at time t=T? would it be at the origin or at $+\frac{A}{2}$ $\endgroup$ Commented Mar 11, 2021 at 15:15
  • $\begingroup$ (a) "Would the time period of the First picture also be T?" I'm not sure what you mean; the time period, $T$, is the time for one cycle, from whatever point in the cycle you start. (b) I'm afraid that I find your diagrams hard to understand. For example, you say that the second diagram is for $t=T/8$, yet a time longer than $T/6$ has elapsed since the first diagram ($t=0$), judging by the change in the position of P. (c) I believe that the answers to all your questions are contained in the equation that I quoted in my answer. I recommend that you use it. $\endgroup$ Commented Mar 11, 2021 at 16:18
  • $\begingroup$ All the diagrams are completely separate and are not related to each other at all. The first diagram depicts an SHM in which the particle begins(t=0) its motion at a displacement of A/2 , the 2nd diagram depicts an SHM in which a particle has its displacement = -A/2 at $t=T/8$ and so on... $\endgroup$ Commented Mar 11, 2021 at 16:51
  • $\begingroup$ I see. Thank you. So perhaps the answer to your question is simply that the particle in the first diagram will not be at $x=-A/2$ at time $T/4$, for the reason that I've explained$. $\endgroup$ Commented Mar 11, 2021 at 16:56

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