AQA M2 Differential Equations

I can see that the particle will slow down so I understand the negative. I don't know why it's x and not x-3.

I've not used I've not used the initial speed nor the 2kg . Does that just come in to part b) which I've not had a go at yet or am I missing something from part a)?

Thanks

Hi

I'm on question 5. The answer I'm getting is not the one in the book

$\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3}$

2nd, corrected, attempt.

The book is wrong, IMO.



It's negative because the force is in the opposite direction to x increasing.



Well you used the 2kg in part a).

The initial speed doesn't come into it at all.

I suspect the question in the book was modified without being checked before they went to press.

Ok. Thanks for the help. Yes, of course I used the 2kg -what was I thinking!


For part b) the answer is
$\frac{d^{2}x}{dt^{2}}= \frac{-10x}{3} \mp 0.1g$
where the sign is + if

and the sign is - if


When the movement is to the right (as in my diagram) the velocity is positive and when the particle is being pulled back the velocity is negative? Does that explain the +/-?

Where does the 0.1g come from or is the book answer all wrapped up in being a modified question?

In part b), you now have to include a frictional force in your equation. If the particle is moving in the positive direction, the friction is acting to the left like the tension. Friction = mu x reaction, reaction = weight and then the 2 will cancel out. Give it a go!

R = mg = 2g
F = muR =0.1x2g

How does the 2 cancel out? I must be in a really thick mode today.

F= 0.2g

Then resultant force is -20x/3 +/- 0.2g and this equals ma = 2a and 2's cancel throughout.

Yep. Got it.

Just to be clear it is -20(x-3)/3 isn't it not -20x/3?