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HSC 2016 MX1 Marathon (archive) (1 Viewer)

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leehuan

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Re: HSC 2016 3U Marathon

Because expecting the unexpected makes the unexpected expected

Q.E.D

Wrong
Would this suffice?

That is a good way to analyse it using physics principles. What I was looking for was something along the lines of sin^2(x) can essentially be written in terms of -cos(2x) and thus we actually walk down to a road of standard derivatives and integrals
 

Nailgun

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Re: HSC 2016 3U Marathon

Wrong

That is a good way to analyse it using physics principles. What I was looking for was something along the lines of sin^2(x) can essentially be written in terms of -cos(2x) and thus we actually walk down to a road of standard derivatives and integrals
h8er detected
 

davidgoes4wce

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Re: HSC 2016 3U Marathon

Think there has been a typo with this solution for part (b) for Circle Geometry



This was the books answer:



I think the answer should be:

















If someone could confirm that would be great.
 

Ambility

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Re: HSC 2016 3U Marathon

Think there has been a typo with this solution for part (b) for Circle Geometry



This was the books answer:



I think the answer should be:

















If someone could confirm that would be great.
I confirm that your working is correct and the book's working is incorrect.
That took some squinting.
 

davidgoes4wce

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Re: HSC 2016 3U Marathon

Can someone again confirm if there is a mistake on this Circle Geometry question?



On that 7th line I believe they have done a mistake. I believe the written expression should be:





If someone could confirm that would be great.
 

leehuan

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Re: HSC 2016 3U Marathon

Expression subject to massive inaccuracy...

 

Jeff_

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Re: HSC 2016 3U Marathon

Could someone please explain why I'm getting a negative answer when I use cosine in the following SHM? When I use sine, the answer is correct.



 

InteGrand

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Re: HSC 2016 3U Marathon

Could someone please explain why I'm getting a negative answer when I use cosine in the following SHM? When I use sine, the answer is correct.



Equating the first method's x to 6 isn't quite right, it should be to -6. This is because by choosing epsilon = pi/2, you made it so the particle starts at the origin but moves towards the negative side first, so you should equate the x to -6. This should then yield the right answer. In fact this can be seen without redoing the Q.: by symmetry of the function (it is odd if you check it), if you equate it to -6, since it attains +6 at t = -1.02, it attains -6 at t = +1.02 (odd function). So the answer is 1.02 (rounded of course).

If you wanted it to be so that it moves towards the positive side first, it should have epsilon = -pi/2 instead.
 

Drongoski

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Re: HSC 2016 3U Marathon

More straightforward this way.


T = 10 ==> n = pi/5

Just choose simplest form:

x(t) = 10 sin (nt) = 10 sin(pi * t/5)

With x = 10 sin(pi *t/5) = 6

pi*t/5 = arcsin(6/10) = 2pi * 36.869. . ./360

.: t = 368.698 . . ./360 = 1.02416 . . . seconds
 
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