Very hard mechanics question (1 Viewer)

James Smith The Third · Sep 19, 2020

I am unsure as to where to start with this question. It is the final question of a 2020 trial paper my friend gave me. Any advice would be appreciated thanks!

ultra908 · Sep 19, 2020

isnt this the grammar q16 lol
which part of the question are you having problem with?

James Smith The Third · Sep 19, 2020

ultra908 said:
isnt this the grammar q16 lol
which part of the question are you having problem with?

I just dont know where to start for even part i. Could you give me some advice please?

Trebla · Sep 19, 2020

For part (i), use differentiation to find the velocity and acceleration equations and do a bunch of substitutions to reach the form of the acceleration equation given

CNSie · Sep 19, 2020

For part (ii) solve $\dot{x} = 0$ .
For part (iii) note that the total distance should be
$\sum_{r=0}^\infty x_r$
where
$x_0 = A$
Get a general formula for $x_r$ (same process as part 2 but using the general solution for tan) and note that your total distance will be in terms of A (constant), and k/n which is a function of P/Q

ultra908 · Sep 20, 2020

CNSie said:
For part (ii) solve $\dot{x} = 0$ .
For part (iii) note that the total distance should be
$\sum_{r=0}^\infty x_r$
where
$x_0 = A$
Get a general formula for $x_r$ (same process as part 2 but using the general solution for tan) and note that your total distance will be in terms of A (constant), and k/n which is a function of P/Q

its should be 2|x_r| (apart from x_0=A), since you need to go down and up, and the distance cant be negative.

CNSie · Sep 20, 2020

ultra908 said:
its should be 2|x_r| (apart from x_0=A), since you need to go down and up, and the distance cant be negative.

Oh yeah true! But the reasoning holds at least haha

Nav123 · Sep 21, 2020

Part ii/iii written out:
In part i you would find:

$\dot{x}=-Ae^{-kt}(n\sin(nt)+k\cos(nt))$
To find when particle is rest set $\dot{x}=0$ which you would do in part ii

$-Ae^{-kt}(n\sin(nt)+k\cos(nt))=0$
$-Ae^{-kt} \neq 0$ so solve:
$(n\sin(nt)+k\cos(nt))=0$
mucking around with auxiliary angle (or you could just divide by cos) we get:
$\sqrt{n^2+k^2}\sin(nt+\alpha)=0 \hspace{3mm}\text{where}\hspace{3mm} \alpha=\arctan(\frac{k}{n})$
$\sqrt{n^2+k^2}\neq 0$ so solve:
$\sin(nt+\alpha)=0$ not sure if general solutions is in the syllabus so I'll just list out the solutions
$t=\frac{\pi-\alpha}{n},\frac{2\pi-\alpha}{n},\frac{3\pi-\alpha}{n},...$
Then sub this into x (keeping in mind the displacement will be negative) you will get the answer for ii.

For part iii like previous posters said take the infinite sum so we get the sum (i'll write it out long assuming general sol isn't in the syllabus):
$D_t=A-2Ae^{-k(\frac{\pi-\alpha}{n})}\cos(\pi-\alpha)+2Ae^{-k(\frac{2\pi-\alpha}{n})}\cos(2\pi-\alpha)-2Ae^{-k(\frac{3\pi-\alpha}{n})}\cos(3\pi-\alpha)+...$
Notice the pattern for odd multiples of $\pi$ cosine is negative and for even multiples of $\pi$ it's positive. So simplifying the sum:
$D_t=A+2Ae^{\frac{k\alpha}{n}}(e^{\frac{-k\pi}{n}}\cos{\alpha}+e^{\frac{-2k\pi}{n}}\cos{\alpha}+e^{\frac{-3k\pi}{n}}\cos{\alpha}+...)$
$D_t=A+2Ae^{\frac{k\alpha}{n}}\cos{\alpha}(e^{\frac{-k\pi}{n}}+e^{\frac{-2k\pi}{n}}+e^{\frac{-3k\pi}{n}}+...)$
We can see the sum is convergent quite easily but a formal proof is prob required. Using the GP sum we finally get:
$D_t=A-2Ae^{\frac{k\alpha}{n}}\cos{\alpha}\left( \frac{e^{\frac{-k\pi}{n}}}{1-e^{\frac{-k\pi}{n}}} \right)$
$D_t=A-2Ae^{\frac{k\alpha}{n}}\cos{\alpha}\left( \frac{1}{e^{\frac{k\pi}{n}}-1} \right)$
$\text{Now we need to evaluate} \hspace{3mm} \cos({\arctan(\frac{k}{n})) \hspace{3mm} \text{drawing up a triangle we find} \hspace{3mm} \cos({\arctan(\frac{k}{n}))=\frac{n}{\sqrt{n^2+k^2}$
$\text{Evaluate this in terms of P and Q we get} \hspace{3mm} \frac{n}{\sqrt{n^2+k^2}}=\sqrt{1-\frac{Q^2}{4P^2}}\hspace{3mm} \text{notice how this depends entirely on} \hspace{3mm} \frac{Q}{P} \hspace{3mm}.$
$\text{The final task is to find an expression in terms of P and Q for the powers of e namely} \hspace{3mm} \frac{k\alpha}{n}, \frac{k\pi}{n}$

$\frac{k\alpha}{n}=\frac{0.5Q}{0.5\sqrt{4P^2-Q^2}}\times \sqrt{\frac{4P^2-Q^2}{4P^2}}=\frac{Q}{2P}$
$\text{since P is postive no need to worry about absolute values. Again depends only on} \hspace{3mm} \frac{Q}{P}$ .
$\frac{k\pi}{n}=\frac{0.5Q\pi}{0.5\sqrt{4P^2-Q^2}}=\frac{\pi}{\sqrt{\frac{4P^2}{Q^2}-1}}$
Therefore every part of the expression for $D_t$ depends solely on the fraction $\frac{Q}{P}.$ (notice that A is constant so we don't have to worry about that).

Sidenote: Sorry for the terrible latex

. I also probably included too many steps which wouldn't be required in a typical exam.

Very hard mechanics question (1 Viewer)

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