Question need help immediately (1 Viewer)

KingOfActing · Jul 5, 2016

Use $\text{Period} = \frac{2\pi}{n}$ to find n, then differentiate and substitute $t = 1$ to get two equations in two variables which you can solve like any other.

hihihi · Jul 6, 2016

been there done that, just need someone to solve the two equations for me plox thx 360noscope bai

InteGrand · Jul 6, 2016

hihihi said:
been there done that, just need someone to solve the two equations for me plox thx 360noscope bai

$\noindent Since the period is $8\pi$, we have $n=\frac{1}{4}$. Note we can find $a$ quickly by using $v^2 = n^2 \left(a^2 -x^2\right)$. Sub. in the given data of $v=-1$ at $x=3$, so $1 =\frac{1}{16}\left(a^2 -9\right)$. Solving for $a$ yields $\boxed{a = 5}$. Now, we know $x = 3$ when $t=1$. So subbing into the $x(t)$ formula, $3 = 5\sin \left(\frac{1}{4}+ \alpha\right)$. So $\boxed{\alpha = \sin^{-1}\left(\frac{3}{5}\right) -\frac{1}{4}}$, which you can use your calculator to find a numerical value for.$

Edit after tywebb's post: sorry, the alpha here is the one that gives v = 1 rather than -1 at t = 1. The other solution for alpha in [0, 2pi) is the desired one, as tywebb has explained below (see his post). So correct alpha (call it X), has

3/5 = sin(1/4 + X), so 1/4 + X = pi – arcsin(3/5). So X = pi – arcsin(3/5) – 1/4.

Thanks go to tywebb for pointing out the mistake.

davidgoes4wce · Jul 6, 2016

InteGrand said:
$\noindent Since the period is $8\pi$, we have $n=\frac{1}{4}$. Note we can find $a$ quickly by using $v^2 = n^2 \left(a^2 -x^2\right)$. Sub. in the given data of $v=-1$ at $x=3$, so $1 =\frac{1}{16}\left(a^2 -9\right)$. Solving for $a$ yields $\boxed{a = 5}$. Now, we know $x = 3$ when $t=1$. So subbing into the $x(t)$ formula, $3 = 5\sin \left(\frac{1}{4}+ \alpha\right)$. So $\boxed{\alpha = \sin^{-1}\left(\frac{3}{5}\right) -\frac{1}{4}}$, which you can use your calculator to find a numerical value for.$

The books answer has got

$a=\frac{32}{\pi} \times \sqrt{2}$

$\alpha=\frac{\pi}{4}$

The answer seemed a bit different to yours.

InteGrand · Jul 6, 2016

davidgoes4wce said:
The books answer has got

$a=\frac{32}{\pi} \times \sqrt{2}$

$\alpha=\frac{\pi}{4}$

The answer seemed a bit different to yours.

$\noindent Are you sure you're looking at the right question's answer? If we assume those, we have $x(t) = \frac{32\sqrt{2}}{\pi} \sin \left(\frac{1}{4}t +\frac{\pi}{4}\right)$. Then the RHS here should equal $3$ at $t=1$, but instead it is $\frac{32\sqrt{2}}{\pi} \sin \left(\frac{\pi +1}{4}\right)$, which isn't equal to 3 (as an exercise, you may want to show this without use of a calculator).$

jathu123 · Jul 6, 2016

yeah InteGrands answer is right

davidgoes4wce · Jul 6, 2016

My bad I was looking at Q 18.

tywebb · Jul 6, 2016

InteGrand said:
$\noindent Since the period is $8\pi$, we have $n=\frac{1}{4}$. Note we can find $a$ quickly by using $v^2 = n^2 \left(a^2 -x^2\right)$. Sub. in the given data of $v=-1$ at $x=3$, so $1 =\frac{1}{16}\left(a^2 -9\right)$. Solving for $a$ yields $\boxed{a = 5}$. Now, we know $x = 3$ when $t=1$. So subbing into the $x(t)$ formula, $3 = 5\sin \left(\frac{1}{4}+ \alpha\right)$. So $\boxed{\alpha = \sin^{-1}\left(\frac{3}{5}\right) -\frac{1}{4}}$, which you can use your calculator to find a numerical value for.$

jathu123 said:
View attachment 33323

yeah InteGrands answer is right

$\text{One of them is and the other isn't.}$

$a=5\text{ but }\alpha = \sin^{-1}\left(\frac{3}{5}\right) -\frac{1}{4} \text{ gives }v=1\text{ not -1. So this value of }\alpha\text{ is wrong.}$

$\text{You should also consider }\alpha =\pi- \sin^{-1}\left(\frac{3}{5}\right) -\frac{1}{4}.\text{ This gives the correct value of }v=-1\text{ and so the textbook answer of }\alpha\doteqdot2.248\text{ is correct.}$

$\text{For the record too this is Cambridge 3 unit year 12 Ex. 3D Q19}$

tywebb · Jul 6, 2016

InteGrand said:
The other solution ... is the desired one.

You mean MY solution is the desired one because mine is correct and yours is wrong. You've gone back and edited your post after reading mine without acknowledging this fact.

If someone reads this thread tomorrow or any time thereafter they will not see it this way so I'm just going to point that out.

InteGrand · Jul 6, 2016

tywebb said:
You mean MY solution is the desired one because mine is correct and yours is wrong. You've gone back and edited your post after reading mine without acknowledging this fact.

If someone reads this thread tomorrow or any time thereafter they will not see it this way so I'm just going to point that out.

Other solution as in there are two solutions in that interval, the one I had isn't the right one, and the one you had is.

Sure, I'll acknowledge your input in the post.

tywebb · Jul 6, 2016

InteGrand said:
Thanks go to tywebb for pointing out the mistake.

edit note

Last edited by InteGrand; Today at 9:02 PM. Reason: Made a mistake, corrected by tywebb below. This post was not edited until after seeing tywebb's post.

That's much better.

Drongoski · Jul 6, 2016

tywebb said:
That's much better.

InteGrand is one of the most selfless, most accomplished maths guys on BoS; never uttered a word in anger or rebuke, always patient, always helpful, never seeking credit for all the great work he has done here. I cannot imagine he would, consciously, seek credit for work done by others without attribution. I'm sure this was just a small oversight on his part.

In days of yore, we'd have called InteGrand "a gentleman and a scholar"

davidgoes4wce · Jul 6, 2016

Drongoski said:
InteGrand is one of the most selfless, most accomplished maths guys on BoS; never a word in anger, always patient, always helpful, never seeking credit for all the great work he has done here. I cannot imagine he would, consciously, seek credit for work done by others without attribution. I'm sure this was just a small oversight on his part.

Just wondering do you and Integrand know each other in real life?

InteGrand · Jul 6, 2016

Yeah I did this Q. in my head in the early morning so I'm not too surprised that there was an error in part of it haha. I edited initially after reading tywebb's post without writing I saw that post because I believed the main purpose of these threads was just to help answer people's Q's (and benefit future readers also), so the main thing on my mind was to just note the correction of the error in the post to prevent future readers' confusion. I wasn't thinking to gain credit or 'steal' tywebb's work or anything, so apologies to tywebb if it came across as such. This was why, upon realising that tywebb wanted me to acknowledge his post and that I'd made a mistake (which I realised from one of his later posts), I was happy to do so (which is why I re-edited the post).

Drongoski · Jul 6, 2016

davidgoes4wce said:
Just wondering do you and Integrand know each other in real life?

Not at all. I haven't the foggiest who he really is.

Green Yoda · Jul 6, 2016

Side note: Also like to acknowledge the million times Integrand has helped us out..you the real mvp man

Drongoski · Jul 7, 2016

Rathin said:
Side note: Also like to acknowledge the million times Integrand has helped us out..you the real mvp man

1000000₂ ??

Green Yoda · Jul 7, 2016

Drongoski said:
1000000₂ ??

The exaggeration was needed to show gratitude.

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