WANTED Solutions to SGS Trials 2004 (1 Viewer)

[mojako]

yes I get part b

To anyone reading:
for this Q, do you think we're allowed to work backwards / use the given result (in an obvious way... not like working backwards but pretending to work forwards all the time)?

mm should I tell Mr Black? Obviously not since he'll delete his post and say that he already got it and give me a "" like he did for part a
and more.. he could be working with Lord Voldermort

 

[nit]

Yeh, Mojako, that's basically what i did in the exam for that q..."intelligent" use of question

It worked out, and i got all the marks, so using the q without assuming it will work, but there are much better ways for this q than plainly bashing it out.

 

[mojako]

>> so using the q without assuming it will work <<
When we did it we assume it will work ^^
if we don't assume it will work, why do we do it?

>> but there are much better ways for this q than plainly bashing it out. <<
Hmm.... like what?

oh, your target UAI.. will you achieve it?

 

[nit]

hmm, i didn't do it the fancy way, so i'll have to dig up the solutions...my own solution was abt 4 pages long, hence ungainly. Once i've found it i'll try to post it up.

As for target UAI, probably not, but it's worth a try

 

[Rorix]

by request
x= asin (nt), T= 2pi/n.
V=ancos(nt)
At P, t(1)= 1/n arcsin D/a.
Therefore V = an cos arcsin D/a
= an sqrt(a^2-d^2)/a
=n sqrt (a^2 - d^2)
i.e. RTP t(2) - t(1) = 2/n arctan sqrt(a^2 - D^2)/D
Now, the second time through P
P=a sin (pi-nt(2))
Rearranging, t(2) = 1/n (pi - arcsin D/a)
Therefore, t2-t1 = 1/n (pi - 2arcsin D/a)
t2-t1 = 2/n (pi/2 - arcsin D/a)
From the triangle used earlier to solve cos arcsin D/a, we have a right angled triangle with hypotenuse a, other sides D and sqrt(a^2 - D^2)
arcsin D/a is the angle between the sides a and sqrt (a^2-D^2)
pi/2 - arcsin D/a is this the angle between sides D and a (as it is a right angled triangle, angle sum)
This angle also equals arctan (opposite/adjacent) = arctan sqrt(a^2 - D^2)/D
Therefore t2 - t1 = 2/n arctan sqrt(a^2-D^2)/D
Now, moving toward the answer just involves reintroducing the terms i cancelled to get the RTP.

 

[mojako]

Rorix.. u gave the answer away to Mr Black!!

 

[spoilt brat]

hey guys!... thanks for all that help!!.. wit the first question that is. i totally get it!!!.. now what's troubling me the most is the nest part... simple harmonic motion.. i have no idea what to do??.. is there any way to solve it without working backwards, from the answer given??... Please let me know!

Thanks in advance

 

[mojako]

Rorix did that without working backwards
if he did work backwards, he didn't write it in the final answer (a.k.a. the writing part of the exam booklet)

Oh, just realised that, he made it invisible.
But it's still there...

 

[spoilt brat]

THANK YOU so much, Rorix and Mojako!!!

 

[Maxsaw]

Pm sent.



Maxsaw.