James Smith The Third
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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!
Very hard mechanics question (1 Viewer)
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James Smith The Third
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I just dont know where to start for even part i. Could you give me some advice please?
CNSie
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For part (ii) solve 
.
For part (iii) note that the total distance should be
where
Get a general formula for
(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
For part (iii) note that the total distance should be
where
Get a general formula for
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.For part (ii) solve
For part (iii) note that the total distance should be
where
Get a general formula for
CNSie
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Oh yeah true! But the reasoning holds at least haha
Part ii/iii written out:
In part i you would find:
+k\cos(nt)))
To find when particle is rest set
which you would do in part ii
+k\cos(nt))=0)
so solve:
+k\cos(nt))=0)
mucking around with auxiliary angle (or you could just divide by cos) we get:
=0 \hspace{3mm}\text{where}\hspace{3mm} \alpha=\arctan(\frac{k}{n}))
so solve:
=0)
not sure if general solutions is in the syllabus so I'll just list out the solutions
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):
}\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
cosine is negative and for even multiples of it's positive. So simplifying the sum:
)
)
We can see the sum is convergent quite easily but a formal proof is prob required. Using the GP sum we finally get:
)
)
) \hspace{3mm} \text{drawing up a triangle we find} \hspace{3mm} \cos({\arctan(\frac{k}{n}))=\frac{n}{\sqrt{n^2+k^2})
.
Therefore every part of the expression for
depends solely on the fraction 
(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.
In part i you would find:
To find when particle is rest set
mucking around with auxiliary angle (or you could just divide by cos) we get:
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):
Notice the pattern for odd multiples of
We can see the sum is convergent quite easily but a formal proof is prob required. Using the GP sum we finally get:
Therefore every part of the expression for
Sidenote: Sorry for the terrible latex
. I also probably included too many steps which wouldn't be required in a typical exam.
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