Hello,
I was wondering if someone could please help me as I am not understanding part c of this question.
A particle moves in a line so that its distance from the origin at time t is x.
b) If a=3x-2x^3 and v=0 at x=1 find v in terms of x.
I have already done this and my answer was v^2=3x^2-x^4-2, which is correct.
c) Describe the motion of the particle. Is it in simple harmonic motion?
I have no idea how to do part c and in the worked solutions I don't understand what they do as they have v^2=y in the solutions.
Thankyou 
Hey! So if we factorise that expression:
(x^2-2)\\v^2=(x-1)(x+1)(x-\sqrt{2})(x+\sqrt{2}))
So the velocity will be zero (meaning the particle is stationary) at 4 points - \(\pm1,\pm\sqrt{2}\). You would describe the motion in that sense - Where it stops.
Let's examine that a little more closely. We know that \(v=0\) at \(x=1\). At this point, the acceleration is positive, so it will proceed to move to the right. By the time it stops again at \(x=\sqrt{2}\), the acceleration is negative, meaning that it will move to the left after this. What is actually going to happen is that the particle will oscillate between \(x=1\) and \(x=\sqrt{2}\), never moving from between these two points. Indeed, if we find the point where acceleration is zero:
=0\\x(\sqrt{3}-\sqrt{2}x)(\sqrt{3}+\sqrt{2}x)=0\\x=0,\pm\frac{\sqrt{3}}{\sqrt{2}})
It is between the two stop points, which makes sense. Buuut, it isn't exactly halfway between, so it is
not simple harmonic motion for that reason
This is how I'd look at it - I think you could also go about converting the function to a function of time, not position, but pretty sure you'd need to do a nasty integral to do that.
What did the solutions do? Does it seem similar to my approach?
(\(v^2=y\) was probably just for convenience)
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