Hey if anyone could solve this integral for me it would be much appreciated:
Note: I have only done 3 unit mathematics (Australian HSC) and I don't start Integral calculus until Semester 2, so there are several integral properties I know nothing about. In fact I'm not sure if this can be solved (looking on wikipedia I was unable to find a formula):

(L^.5)dr
________________
L((mgcos(r/L))^.5)

This has the limits and the integration sign left out. I have left out the limits because I'm still working on defining them so if anyone who has the time to solve this can leave the integrated function in the form [f(x)] I'll know that I need to sub the limits in.

If the above is a little difficult to read, once constants (L, m and g) are taken out the integral is

dr/((cos(r/L))^.5) with the denominator being the square root of cos(r/L).

Thanks

2. Originally Posted by behemoth100
Hey if anyone could solve this integral for me it would be much appreciated:
Note: I have only done 3 unit mathematics (Australian HSC) and I don't start Integral calculus until Semester 2, so there are several integral properties I know nothing about. In fact I'm not sure if this can be solved (looking on wikipedia I was unable to find a formula):

(L^.5)dr
________________
L((mgcos(r/L))^.5)

This has the limits and the integration sign left out. I have left out the limits because I'm still working on defining them so if anyone who has the time to solve this can leave the integrated function in the form [f(x)] I'll know that I need to sub the limits in.

If the above is a little difficult to read, once constants (L, m and g) are taken out the integral is

dr/((cos(r/L))^.5) with the denominator being the square root of cos(r/L).

Thanks
$\int dr \, \frac{\sqrt{L}}{L \sqrt{mgr \, cos \left ( \frac{r}{L} \right ) }}$

Let $\theta = \frac{r}{L}$. Then $d \theta = \frac{dr}{L}$. So...

$\int dr \, \frac{\sqrt{L}}{L \sqrt{mgr \, cos \left ( \frac{r}{L} \right ) }} = \int d \theta L \cdot \, \frac{\sqrt{L}}{L \sqrt{mg \theta L \, cos( \theta) }}$ = $\frac{1}{\sqrt{mg}} \int \frac{d \theta}{\sqrt{\theta \, cos( \theta )}}$

I can't think of a way to integrate this.

Where did this integral come from anyway?

-Dan

3. Sorry I must not have been clear. The denominator is simply L(mgcos(r/L) to the power of a half) i.e. in your last step there is no theta outside of the cos(theta).
To answer your question I...uh ... well I figured the lecturers were lying to us about the motion of a pendulum being simple harmonic even at small theta so I endevoured to come up with a proof to show the real equation for period lol.
Anyways I guess 10 weeks into advanced physics isn't long enough to be arrogant enough to think you can come up with an equation of periodic motion for a pendulum huh?
Anyways I ended up with that integral and I'm really hoping to get a final equation, even if its incorrect (I'm a tad stubborn). However if this integral is unsolvable chances are I went wrong somewhere huh.
Oh and I'm really hoping that you can integrate it in terms of dr and not dtheta as I spent a while converting it into that form (as it involved integrating a function of theta by dt so I had to find an alternate form of dt).
If you are interested in my method I'll be happy to tell you (not sure if its sound though; as I said only 10 weeks in ) still I would really appreciate it if you were able to integrate the function anyway (in terms or dr )
Edit: I know it would be a simple matter to switch it back after the dtheta integral but I'd like to see the working out for dr thats all. However since your the one helping me, then whatever works for you

4. Originally Posted by behemoth100
Hey if anyone could solve this integral for me it would be much appreciated:
Note: I have only done 3 unit mathematics (Australian HSC) and I don't start Integral calculus until Semester 2, so there are several integral properties I know nothing about. In fact I'm not sure if this can be solved (looking on wikipedia I was unable to find a formula):

(L^.5)dr
________________
L((mgcos(r/L))^.5)

This has the limits and the integration sign left out. I have left out the limits because I'm still working on defining them so if anyone who has the time to solve this can leave the integrated function in the form [f(x)] I'll know that I need to sub the limits in.

If the above is a little difficult to read, once constants (L, m and g) are taken out the integral is

dr/((cos(r/L))^.5) with the denominator being the square root of cos(r/L).

Thanks
It does have a closed form in terms of a hypergeomentric function and
elementary functions, but that can be of no interest to you as yet.

But if you want to see what Mathematica gives try it on the QuickMath integrator.

RonL

RonL

5. Originally Posted by behemoth100
Sorry I must not have been clear. The denominator is simply L(mgcos(r/L) to the power of a half) i.e. in your last step there is no theta outside of the cos(theta).
To answer your question I...uh ... well I figured the lecturers were lying to us about the motion of a pendulum being simple harmonic even at small theta so I endevoured to come up with a proof to show the real equation for period lol.
Anyways I guess 10 weeks into advanced physics isn't long enough to be arrogant enough to think you can come up with an equation of periodic motion for a pendulum huh?
Anyways I ended up with that integral and I'm really hoping to get a final equation, even if its incorrect (I'm a tad stubborn). However if this integral is unsolvable chances are I went wrong somewhere huh.
Oh and I'm really hoping that you can integrate it in terms of dr and not dtheta as I spent a while converting it into that form (as it involved integrating a function of theta by dt so I had to find an alternate form of dt).
If you are interested in my method I'll be happy to tell you (not sure if its sound though; as I said only 10 weeks in ) still I would really appreciate it if you were able to integrate the function anyway (in terms or dr )
Edit: I know it would be a simple matter to switch it back after the dtheta integral but I'd like to see the working out for dr thats all. However since your the one helping me, then whatever works for you
The pendulum equation works out to be:
$\theta^{ \prime \prime } + \frac{g}{L} sin( \theta ) = 0$
where L is the length of the pendulum and $\theta$ is the angle the string makes with the vertical.

The simple harmonic oscillator equation is:
$x^{ \prime \prime } + \omega ^2 x = 0$

If the initial angular displacement of the pendulum is small enough we may approximate
$sin( \theta ) \approx \theta$

and the pendulum equation becomes the simple harmonic oscillator equation. (So even a simple pendulum really doesn't even follow simple harmonic motion.)

The pendulum equation can be solved in terms of elliptic integrals, which can only be approximated. I haven't seen it done, but CaptainBlack is likely correct that we can write the solution in terms of hypergeometric functions. (At the very least the solution can certainly be expanded into a series of hypergeometric functions.) In Physics the usual method is to use elliptic integrals and expand an approximation from there.

-Dan

6. Yeah we did actually prove the period of a pendulum undergoing simple harmonic motion at small angles where sin(theta)~theta. We then proved it under different kinds of damping, when it has a driving force at natural frequency, and then started all over again with a physical pendulum.

The reason I started all this nonsense was I was a little bored and I wanted to prove the period of the pendulum without using assumptions such as it moves in SHM such that omega = (k/m)^(1/2) or f = (omega/2pi).
As you can see I failed abysmally.

Anyway I thank you both for taking the time to help me out and have just one more request. I haven't done hypergeometric functions (as captainblack guessed) so I can't understand the intergrated function.

Therefore I was wondering if the integral can be represented in terms of elementary functions only if I gave you the limits ((theta)L, zero) or if real numbered constants are required (pi/18),0 (where theta = 10 degrees in radians and L = 1 metre).

The integral is multiplied by the constant 4.

Thanks

7. Actually I just remember Captainblack's calculator so I plugged it into that, and came up with... a negative period.

Fantastic

8. Originally Posted by topsquark
The pendulum equation works out to be:
$\theta^{ \prime \prime } + \frac{g}{L} sin( \theta ) = 0$
where L is the length of the pendulum and $\theta$ is the angle the string makes with the vertical.
Does the differencial equation for the pendulum have a solution in terms of elliptic integrals?

Bucause I think it does. For the most part it is just not necessay.

9. Originally Posted by ThePerfectHacker
Does the differencial equation for the pendulum have a solution in terms of elliptic integrals?

Bucause I think it does. For the most part it is just not necessay.
huh, well I am waaaaaaaay out of my league here but looking at my notes it shows that we did prove the period of a pendulum where sin(theta)~ theta and showed via an infinte series that it holds for a maximum of 15 degrees (the infinite series being 1 + ((1^2)/(2^2)) sin^2 (CAPITAL THETA)/2 + ((1^2).(3^2))/((2^2). (4^2))sin^4 (CAPITAL THETA)/2 + ...) (or rather for capital theta = 15degrees the true period is longer than that given by the approximate by less than .05%.
And it is here that my insubstantial contributions end, until I've done about 3 more years of intensive maths .

By the way how do you guys right out your working neatly on these web pages? I have to use ^ for power of and you guys have integrals and stuff.
I clicked on one of the equations you had and it came up with a bunch of code.

10. Originally Posted by ThePerfectHacker
Does the differencial equation for the pendulum have a solution in terms of elliptic integrals?

Bucause I think it does. For the most part it is just not necessay.
Yes, the problem can be solved "exactly" using elliptic integrals. We can then use an approximation of the integral to form approximate solutions to the pendulum equation. Obviously you can approximate solutions to the differential equation without going through all that. I have never heard why the elliptic integral version is used instead of one of these. (Unless it was merely a good excuse to teach us what they are...)

-Dan