# Simple harmonic oscillation

1. Jan 19, 2013

### Ezio3.1415

Trust me this is not homework... My last two questions were removed cause they looked like homework... I understand its the forum policy... From now on I will post the 'seemingly homework' on the homework sections...

Suppose,there's a rod of mass m1 hanging from a point... And a mass m2 is attached to the other end... If I let it oscillate will it present a simple harmonic oscillation... If it does,what will be the period...

My intuition said no...(well the pendulum's string always has to be mass less)
But when I thought mathematically, what's the difference if I find the center of mass of (m1+m2) & solve it... That would only change the L... And I can find the differential equation
by τ=Iα... thinking the center of the mass of m1+m2 acts at a distance L... and I would find the moment of inertia of the system of m1+m2... so what do u think?

2. Jan 19, 2013

### VantagePoint72

Your idea to think about the moment of inertia of the pendulum is a good one. The rotational kinetic energy of a cylindrically symmetric object rotating around an axis through its end is $K=\frac{1}{2}I\omega^2 = \frac{1}{2}I\dot{\phi}^2$ where $\phi$ is the angle between the vertical and the pendulum's central axis (with the over dot meaning the time derivative) and I is the moment of inertia around the end. If the mass of the pendulum is $m$ and the distance from the pivot to the centre of mass is $l$, then the gravitational potential energy of the pendulum is $U=mgl(1-\cos\phi) \approx \frac{1}{2}mgl\phi^2$ where the approximation is valid for small angles of oscillation. Thus, in the small angle approximation, we have the Lagrangian:
$L = K - U = \frac{1}{2}I\dot{\phi}^2 - \frac{1}{2}mgl\phi^2$

The universal identifier for simple harmonic motion in some coordinate is a Lagrangian which is quadratic in that coordinate and its derivative, which we have above for the $\phi$ coordinate. This comes straight from the Euler-Lagrange equations, since $L=\frac{1}{2}(A\dot{q}^2 - Bq^2)$ (for arbitrary constants A and B, with a factor of a half taken out for computational convenience) implies:
$A\frac{d^2q}{dt^2} + Bq = 0$
This is the differential equation for simple harmonic motion with frequency $\omega = \sqrt{\frac{B}{A}}$. So, you can read off the harmonic frequency directly from the Lagrangian for the pendulum: $\omega = \sqrt{\frac{mgl}{I}}$. This is related to the period of oscillation in the usual way.

The general case for a less symmetric pendulum follows similarly using the principle moments of inertia about the centre of mass (which you can either use to determine the moments around the end using the parallel axis theorem, or else just separate the motion into rotation about the centre of mass and the motion of centre of mass itself). This is done in the classical mechanics text by Landau and Lifshiftz in the example problem 3 of section 32, if you want more details.

Last edited: Jan 19, 2013
3. Jan 20, 2013

### Ezio3.1415

Thank you for your answer... Btw, what was the period u found for this problem?
Why our simple pendulum's string always is said to be massless... It provides harmonic oscillation even if it has mass...

4. Jan 20, 2013

### VantagePoint72

I told you the angular frequency, it is very easy to work out the period from that.

Because the analysis is simpler if the string is massless, and the approximation is close enough to work well.

5. Jan 20, 2013

### Ezio3.1415

U used m... Are u referring to the center of mass of m1+m2 at L distance?

6. Jan 20, 2013

### VantagePoint72

I treated the general case of a pendulum of any shape, as long as it's cylindrically symmetric. For your special case, you have to work out the appropriate moment of inertia.

7. Jan 20, 2013

### Ezio3.1415

I found
ω^2=2g(m1+m2)/l(m1+2m2)

Can u please check my calculation?

8. Jan 21, 2013

### sophiecentaur

I can't see where the symmetry is necessary. The MI is whatever it is and the displacement of the CM in either sense will provide the same restoring force if the position of the pivot is not changed, won't it? Isn't that all that's necessary for SHM?

9. Jan 21, 2013

### VantagePoint72

Cylindrical symmetric is necessary for the specific formula I worked out, which is what the sentence you're quoting is referring to. You're right that there is still SHM in the non-symmetric case—the reference I gave works that out explicitly. The point is that in the general case, the motion of the body needs to be decomposed into rotation around each of the three principal axes of inertia and so rotational kinetic energy has three terms. So, the formula I gave needs to be replaced with something more general (given in L&L) to get the angular frequency of a non-symmetric pendulum.

10. Jan 21, 2013

### sophiecentaur

I was thinking in terms of a two dimensional object swinging around one axis - which is a 'trivial' case, I guess. Anything more would get a lot harder. If you had a three dimensional object hanging from a universal bearing then it could have different MIs about different axes and it could also be spinning about another axis. Could you not, still, break this down into three motions, though - two with a restoring force towards the vertical (giving two different periods) and one rotation at a constant rate around an axis between the pivot and the CM? Perhaps there's something more.

11. Jan 21, 2013

### VantagePoint72

The case I was describing is not a pendulum attached to a free bearing. It's still attached on a fixed axis, which will, in general, be at some angle to each of the principal axes of inertia. If the pivoting axis is at an angle of $\alpha$, $\beta$, and $\gamma$ to the principal axes, about which moments of inertia are $I_1$, $I_2$, and $I_3$ (through the centre of mass), then the rotational kinetic energy is $K = \frac{1}{2}I_1\omega^2 \cos\alpha+ \frac{1}{2}I_2\omega^2\cos\beta + \frac{1}{2}I_3\omega^2\cos\gamma$. There is an additional contribution to the kinetic energy from the centre of mass motion (since the principal moments of inertia are taken through the CM, not the pivot point), and the potential energy stays the same. Without looking into it more deeply, it sounds like your idea for how to treat the free bearing case would work, but some of the details sound very messy. As I recall from classical mechanics, a symmetrical top fixed by a bearing to a table top has a fairly complicated description. A non-symmetrical fixed top might be intractable—but classical mechanics was a long time ago...