Scenario 1:

$x = A \sin(\omega t+\delta)$

Now, if we increase t by $\frac{2\pi}{\omega}$, then the displacement becomes

$x' = A \sin(w(t+(2\pi/\omega))+\delta)$

$=A \sin(\omega t + 2\pi + \delta)$

$=A \sin (\omega t+\delta)$


So, $2\pi/\omega$ is the time period. I understand this conclusion completely. I have no problem with this scenario. I only stated scenario 1 to explain the premises on which scenario 2 is based.

Scenario 2:

$x=A \sin(\omega t+s)$

Now if we increase t by $\pi/\omega$ or half of the time period, we get

$x' = A \sin(\omega(t+(\pi/\omega))+\delta)$

$=A \sin(\omega t + \pi + \delta)$

$= - A \sin(\omega t+\delta)$

$= -x$

Now, I don't understand this conclusion. I think that the displacement $x'$ should've been equal to $x$, not $-x$.

You can visualize this in your head. Suppose we have a particle at the displacement $x=0.5m$. Let the time period is T. Now after complete oscillation or after time T, no doubt the particle would return to position $x = 0.5m$. But this is true for a half oscillation as well. If the amplitude is A, then we can visualize the particle moving from $x=0.5m$ to $x=+A$ by going right and then back to $x=0.5m$ by going left. There you go! That's a half oscillation. And after half oscillation, what do we see? We see that $x'=x$, contrary to the above mathematical conclusion where $x'=-x$. So is the above mathematical conclusion wrong?


2 Answers 2


Your reasoning is not quite correct, where the particle will find itself after half a period depends on its initial position.

If you start at $x=0$ then after half a period the oscillator will return to its initial position, i.e. $$x(\pi/\omega) = -x(0) = 0.$$ If you start at the point of maximal displacement, $x=A$ then after half a period the oscillator will find itself at the opposite point of the maximal displacement $$x(\pi/\omega) = -A = -x(0).$$ The rest of the cases fall in between, but as the math has correctly shown, $$x(t+\pi/\omega) = -x(t).$$


The visualization in your head is wrong, assuming your 0.5mrecurring points on sin


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