# Simple Harmonic Motion of a pendulum

1. Dec 1, 2007

### Thomas_

Hello,

I have problems solving the following two problems:
1)You measure the period of a physical pendulum about one pivot point to be T. Then you find another pivot point on the opposite side of the center of mass that gives the same period. The two points are separated by a distance L. Use the parallel-axis-theorem to show that $$g = L(2\pi/T)^2$$

2)A slender, uniform, metal rod with mass M is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring with force constant k is attached to the lower end of the rod, with the other end of the spring attached to a rigid support. If the rod is displaced by a small angle $$\theta$$ from the vertical and released, show that it moves in angular SHM and calculate the period (Hint: Assume that the angle is small enough so that $$sin(\theta) =~ \theta$$and $$cos(\theta) =~ 1$$. The motion is simple harmonic if $$d^2\theta/dt^2$$ = $$-\omega^2\theta$$

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For 1)
The period for a physical Pendulum around P is given by:

$$T=2\pi \sqrt{\frac{I_p}{MgL}}$$

The Parallel axis theorem is:

$$I_p = I_{cm} + Md^2$$ (d is distance from the center of mass).

As far as I understand the problem, L = 2d. However, I can't seem to figure it out. Solving for g gives me:

$$g = (\frac{2\pi}{T})^2 \frac{I_p}{ML}$$

I don't know how $$\frac{I_p}{ML}$$ is supposed to become L.

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For 2) I don't know where to start here. I can't even imagine how this rod would undergo any kind of harmonic motion. The picture confuses me and I can't imagine the rod moving.

2. Dec 2, 2007

### Staff: Mentor

In the second problem, the spring imparts a force kx to the end of the rod, where k is the spring constant and x is the displacement of the traveling end of the spring (and end of rod) from the equilibrium position (zero spring force).

x is also L/2 sin $\theta$, where L is the length of the rod.

If the angle is very small, we approximate $$sin(\theta) =~ \theta$$ in order to make a linear differential equation.

Last edited: Dec 2, 2007
3. Dec 2, 2007

### Thomas_

Hm, thank you.

I used $$\sum\tau = I\alpha)$$
and came up with:
$$\sum\tau = -k\frac{L}{2}sin(\theta)(\frac{L}{2}cos(\theta)) = -k\frac{L^2}{4}\theta$$

$$=> I\alpha + k\frac{L^2}{4}\theta = I\frac{d^2\theta}{dt^2} + k\frac{L^2}{4}\theta = 0$$
That should be SHM.

However, how do they get to $$\frac{d^2\theta}{dt^2} = -\omega^2\theta$$ ?

Also, any ideas for Problem 1)?

4. Dec 2, 2007

### Staff: Mentor

Well - in the second problem, $\omega^2$ would = k/m, and 1/m is related to L2/I.

I seen in the first problem that g is an acceleration based on units of L/T2, so is there a way to relate angular acceleration (which is the same for any point in a rigid body) to that of the linear acceleration, which varies with the distance from the pivot to the location of interest?

5. Dec 2, 2007

### Thomas_

I can't follow you on this one. Yes, I know that $$a_{tan}=r\alpha$$ for a rigid body, but I don't see any angular/linear accelerations in the equation we would have to "convert". My goal is to get $$g=(\frac{2\pi}{T})^2L$$