# Why is the displacement negative when half oscillation is done in simple harmonic motion?

Scenario 1:

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

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

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

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

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

$$=x$$

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?

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, $$x(t+\pi/\omega) = -x(t).$$