Currently in Mathematics we are learning about "Simple Harmonic Motion" and how it satisfies the differential equation $\ddot x = -n^2 x$. Does this mean that the velocity($v$) in terms of displacement($x$) will always have a $\pm$ sign as the velocity will be both positive and negative at different points in space?
My confusion on the matter is further exacerbated by 2 solutions to similar questions (the bolded part of the question is the difference between the two);
Question 1 "A particle moves in a straight line so that at time $t$ its displacement from a fixed origin is $x$ and its velocity is $v$. If the acceleration is $3-2x$, find $v$ in terms of $x$ given that $v=2$ when $x =1$."
Solution: They use the fact that $\frac{d}{dx} (\frac {1}{2} v^2) = \ddot x$ (acceleration) and proceed to solve for $v$, though when they reach $v(x) = \pm \sqrt{6x-2x^2}$, they use the initial conditions to conclude that is must be the positive square root. i.e. $v(x) = + \sqrt{6x-2x^2}$.
Question 2 "A particle of mass 2kg moves in a straight line so that at time $t$ seconds its displacement from a fixed origin is $x$ metres and its velocity is $v$ m/s. If the resultant force (in newtons) that acts on the particle is; $6-4x$, find v in terms of $x$ given that $v=2$ when $x =1$."
Solution: They write $F=ma$, rearrange and conclude that acceleration is $\ddot x = 3-2x$. They then repeat the process (as above) and arrive at $v(x) = \pm \sqrt{6x-2x^2}$, BUT say how that its simple harmonic motion as $\ddot x = 3-2x$, and thus its plus and minus.
Is my understanding of the differential equation and simple harmonic motion correct? If so, is the second solution the correct one?
Thanks
(I'm not sure if this falls more under mathematics or physics, but as I did this in maths, I've put it here)
Best Answer
I tried to sketch a simple harmonic motion below
At the top (a) the mass is moving towards the left, in this case $v<0$. Eventually it will reach the minimum value of $x$, stop, and start moving toward the right. It will get to position (b) where the location is the same as in the previous case, but the velocity has the oposite sign $v >0$. That is, for the same location $x$, the velocity could be have either positive or negative sign. As the mass keeps moving, it will get to the other side of $x = 0$ (c), in this case both $x>0$ and $v>0$. Once again it will stop at its maximum distance from the origin, change its velocity and start moving towards the left, reaching again the same position as before, but with a different sign in the velocity. I think this should answer your first question
In general for a force of the form $$F = -m\omega^2 x$$ the velocity is
$$ v^2 = \pm \left[2(E - \omega^2 x^2) \right]^{1/2} $$
for some integration constant $E$. For another point of view, the solution to $F = m\ddot{x}$ is
$$ x(t) = A\cos(\omega t + \phi) ~ v(t) = -A\omega \sin(\omega t + \phi) $$
where the constants $A$ and $\phi$ are integration constants. As a matter of fact $E = \omega^2 A^2/2$. Due to the oscillatory nature of the $\sin$ and $\cos$ functions, the sign of the position and velocity changes periodically.
This should answer your second question as well, if you write, for exmaple, $v = +\sqrt{(\cdots)}$, you must be certain in which part of the motion cycle. In general you should use $\pm$