Simple Harmonic Motion/Period of a Physical Pendulum

[Kusu]

I'm studying the motion of a physical pendulum, could someone help me make the final step in figuring out how to find the period so I can make predictions before carrying out a practical? Basically I have a meter rule with holes drilled along the length and will be pivoting it at various points.

I've done lots of reading and watched lots of videos, I know simple calculus, I understand moment of inertia and torque, small angle approximations and parallel axis theory. I've done dimensional analysis which resulted in:

$$T=\sqrt{\frac{I}{mgb}}$$

where T = Period, I = Moment of inertia, m = Mass and b = distance from center of mass to pivot.

I know dimensional analysis doesn't give any constants, and I've seen in books that the actual equation is

$$T=2\pi\sqrt{\frac{I}{mgb}}$$

but I'd like to be able to prove this. I know it has to do with angular acceleration and angular frequency, but not sure how. Can anyone point me in the right direction? Most the videos I watch just make the jump and say "we've previously shown why... $$\omega^2=\sqrt{\frac{mgb}{I}}$$ but I can't find any previous videos that show! :D
I think it's the $$\omega^2$$ that I can't see where it comes from.

Any help much appreciated!

K(p.s. If the Latex is messed up I'm probably working on fixing it, I've not used it much)

 

[Buffu]

You made a typo it should be $\omega$ not $\omega^2$.

 

[mpresic]

The easiest derivations of this result requires you to be able to solve a second order differential equation.
Another method using the conservation of energy requires you to solve a first - order differential equation, or equivalently solving a trigonometric integral like integral dx over square root of (a squared - x squared) with a constant. This is usually attacked about second semester calculus.

 

[Kusu]

I've seen it written $$\omega^2$$ in several places though, what makes you say that? Here's a screen clipping of a video from (credit to Dan Fullerton on Youtube)

I don't know where $$\theta=A cos(\omega t)$$ comes from either, but heard there are two uses for $$\omega$$ in rotional motion?

 

[Kusu]

Ah, I'm not at university. I know integration and differentiation, but don't know how to solve equations, I'll turn my attention there, thank you!

 

[mpresic]

In general torque = moment of inertia times the angular acceleration = I alpha = I d / dt angular velocity = I second derivative of theta with respect to time.
This torque is also equal to Force ( - mg ) times moment arm (b sin theta). Now for small angles sin theta is approximated by theta, so the torque is -mgb theta.
equating the torques gives a second order differential equation for theta. I times second derivative (d 2 theta / d t squared) = - m g b times theta

Now most text "guess" the solution theta = A cos w t. (w is the frequency)

Then they show the second derivative of this solution using calculus is : d 2 theta / d t squared = - A w squared time theta.
we also have derivative (d 2 theta / d t squared) = - (m g b / I ) times theta
This allows us to make the identification of w squared with mgb / I

 

[Kusu]

Thanks for all your input, I think I need to learn how to solve differential equations, or at least how to prove something is the solution of one. Probably the same thing :)

Just to clear up:
$$\tau=I\alpha$$
$$\tau=I\frac{d}{dt}\mbox{angular velocity (sometimes }\omega\mbox{)}$$
$$\tau=I\frac{d^2\theta}{dt^2}$$
Because of small angle approximations we can get rid of 'sin' in the calculation of the torque cased by mg:
$$\tau=-mgb\theta$$
Equating the torques gives
$$-mgb\theta=I\frac{d^2\theta}{dt^2}$$
rearranged to
$$\frac{d^2\theta}{dt^2}+\frac{mgb}{I}\theta =0$$

So what I need to do is learn how to prove that solving the above gives
$$\theta =A\cos \ \omega \ t\mbox{ (this time }\omega\mbox{ is frequency)}$$
and then learn how to show the second derivative of that is
$$\frac{d^2\theta}{dt^2}=-\frac{mgb}{I}\theta$$
which I can do from reading textbooks/videos?

(again, if the Latex is screwy I'm working on it :D )

 

[Buffu]

I've seen it written $$\omega^2$$ in several places though, what makes you say that? Here's a screen clipping of a video from (credit to Dan Fullerton on Youtube)

View attachment 203485I don't know where $$\theta=A cos(\omega t)$$ comes from either, but heard there are two uses for $$\omega$$ in rotional motion?

Is $\omega^2=\sqrt{\dfrac{mgb}{I}}$ correct ?

Sorry I don't know much.

 

[Andrea Vironda]

Is $\omega^2=\sqrt{\dfrac{mgb}{I}}$ correct ?

Sorry I don't know much.

it is $\omega=\sqrt{\dfrac{mgb}{I}}$, that's because a general $\alpha(t)=-\omega^2*x(t)$

 

[kith]

If you are specifically interested in where the factor of $2\pi$ comes into play, have a look at a simple harmonic motion where the displacement $s(t)$ is described by a sine function.

$T$ is the period of the motion, so we need to have $s(T) = s(0)$. Since the period of the sine is $2\pi$, this number needs to be involved somehow. Now check that the following ansatz
$$s (t) = A \sin\left(\frac{t}{T} 2\pi\right)$$
has the desired properties: as $t$ goes from $0$ to $T$, the argument for the sine goes from $0$ to $2\pi$ (sketching the function may help).

From this, you see that $\omega = \frac{2\pi}{T}$.

 

[Kusu]

Thanks to everyone that commented, it's starting to make much more sense now I've been reading up and done some kitchen practicals! Think I'll have a nice grasp soon. Buffu, I see what you mean with my typo now in post #1, I'd mixed the two equations together, even the screenshot I posted in post #3 showed I was wrong! Kith and mpresic you've been really helpful.

We had a lot of fun doing the practicals and we're now trying to get a practical as close as possible to the prediction. Materials and fabrication have helped a lot :) , but could someone help me with error calculations? I watched a Walter Lewin video where he talks about a simple pendulum, and the uncertainty in the length being 1%, and once the root is taken it becomes 0.5%. My equations is much more complicated, I have simplified $$T=\sqrt{\frac{I}{mgb}}$$ to cancel the mass, and include the parallel axis theorem, giving $$T=\sqrt{\frac{L^2+12b^2}{12gb}}$$

Is there a simple explanation as to what to do with my percentage errors in measurement? I have the uncertainty of L to be 0.1% and b varies so will need to be calculated at each pivot point, but I'm sure will follow 1 rule for any position?

 

[kith]

There's the general formula and many special cases. A superb text about this is "An introduction to error analysis" by Taylor. Or search for things like "gaussian error propagation" or "error analysis introductory lab".