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The Wave Function: Solving Equations - HSN forum

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The Wave Function: Solving Equations


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#1 Adam

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Posted 27 April 2006 - 06:19 PM

root3cosx + sinx = root2

therefore 2cos(x - pi.gif/6) = root2

cos(x - pi.gif/6) = root2/2 = 1/root2

(x - pi.gif/6) = pi.gif/4 or 7IPB Image/4


My question is, where does the 7IPB Image/4 bit come from? I understand where the IPB Image/4 comes from, but not 7IPB Image/4. Can somebody please explain? smile.gif

Thank-you.

#2 Mr H

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Posted 27 April 2006 - 07:15 PM

From the CAST rule.

45 degree.gif is the acute angle.

Use the CAST diagram to find the second solution for positive cos.

i.e. in the 4th quadrant.

So 2nd solution (in degrees for easier illustration) is 360 degree.gif - 45 degree.gif = 315 degree.gif.

Which is 7 x 45 degree.gif, in other words 7 pi.gif /4.

H tends 2 infinity

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#3 Adam

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Posted 27 April 2006 - 08:47 PM

Hmm.. I think I understand. But what about this one?

cosx - root3sinx = 1

k = 2
alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5
-> x - 300 = 60, 300?


(the final answers in this question are 0, 4IPB Image/3 and 2IPB Image :s)

#4 ad absurdum

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Posted 27 April 2006 - 09:12 PM

QUOTE(Betinho @ Apr 27 2006, 09:47 PM) View Post

Hmm.. I think I understand. But what about this one?

cosx - root3sinx = 1

k = 2
alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5
-> x - 300 = 60, 300?

I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Say your looking for values of x that satisfy this equation with the range:


\begin{align*}0 \le x \le 360\end{align*}

Now we have:


\begin{align*}cos(x - 300) = 0.5   ,   \end{align*}


\begin{align*}x - 300 = cos^{-1}(0.5)   ,   \end{align*}


\begin{align*}x = cos^{-1}(0.5) + 300   ,   \end{align*}

Now if we took the normal values for the inverse cosine of 0.5 we would get the angle of 60o, and we would say that within the range the values are 60o and 300o
However, we are adding 300o onto the angle, and if we add 300o onto 300o then we get a value outwith our range.
However, adding 300o onto 60o gives us a value still within the range so it's fine to use this
There is another value for x though, and if you think of this as coming from the cosine graph before zero (because it repeats, just as there is a repeat of the graph after 360 there is one before zero). So if we took -60 (i.e. the equivalent of the quadrant four angle in the graph before zero) as one of our values for x, added 300 onto this and got 240 this would give us a value still within our range. Also, if we took the quadrant one angle from our graph before zero, which is -300 (i.e. -360 + 60), and add 300 onto this we get zero which is still in our range.
So from this we get the results:
x = 0, 240, 360

That's how I do it anyway, not sure about how good a method it is
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#5 Mr H

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Posted 27 April 2006 - 09:16 PM

Beat me to it ad adsurdum!

Too slow at typing!!

Following from your last line Betinho;

x = 360 degree.gif, 600 degree.gif

As ad adsurdum states, 600 degree.gif is outwith the usual 0 < x < 360 domain.

The period/wavelength is 360 degree.gif as there is only a single 'x' therefore every 360 degree.gif there will be another solution.

The 360 degree.gif solution above is OK, there will also be a solution 360 degree.gif before it, i.e. 0 degree.gif, and also 360 degree.gif before 600 degree.gif i.e. 240 degree.gif.

So the solutions are (in degrees) 0 degree.gif, 240 degree.gif, 360 degree.gif

Or 0, 4 pi.gif/3, 2 pi.gif in radians.

H tends 2 infinity

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#6 Mr H

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Posted 27 April 2006 - 09:27 PM

QUOTE(ad absurdum @ Apr 27 2006, 10:12 PM) View Post
I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Your method is fine ad adsurdum!

H tends 2 infinity

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#7 Adam

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Posted 27 April 2006 - 09:27 PM

Thanks guys!

It makes sense now. It was beginning to annoy me - I've been staring at that question for days. smile.gif

#8 ad absurdum

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Posted 27 April 2006 - 09:49 PM

QUOTE(Mr H @ Apr 27 2006, 10:27 PM) View Post

QUOTE(ad absurdum @ Apr 27 2006, 10:12 PM) View Post
I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Your method is fine ad adsurdum!
Ah good to hear, I was worried that there was a quicker way I was failing to notice smile.gif
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