Does a Meter Stick Oscillate Faster Than a Simple Pendulum?

[Lalasushi]

confused! about period of pendulum

hey guys, i have a question. "A meter stick oscillates back and forth about a pivot point at one of its ends. Is the period of a simple pendulum of length 1m greater than, less than or the same as the period of the meter stick?"

i thought that the period wouldn't depend on the mass because of what Galileo Galilei found (that the period doesn't change with the amplitude or the mass) so...which i found confusing because the formula does not contain m in it (T = 2 pi root L/g) and the correct answer is "greater than"...i found on some websites that actually the amplitude does change slightly...but what about the mass?? can someone please explain?

 

[dextercioby]

There's a difference between the physical and the mathematical pendulum. The first is a generalization of the second and the formula of the period for small tautochrone oscillations for both is

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

, where "I" is the moment of inertia of the oscillating body of mass "m" wrt to an axis perpendicular to the plane of oscillation passing through the point of suspension, and "d" is the distance between the CM of the oscillating body and the point of suspension.

Daniel.

 

[Jelfish]

For small oscillations, the angular frequency of a pendulum is:

$$\omega = \sqrt{\frac{g}{L}}$$

where g is acc. of gravity and L is the length to the center of mass. The conclusion that Galileo found was that for said small oscillations, the mass of the pendulum had no bearing to the oscillation period, only the length.

So for your question, consider what "L" is for both situations. For the mass on a string, it's usually assumed that the string has negligible mass, so the length would be 1m. However, for the meter stick, the mass is not concentrated at the end. In fact, it's concentrated at the center of mass, which, if the meter stick has equal mass density throughout, should be in the middle, i.e. 0.5m. Since the meter stick has a smaller "L," then the angular frequency will be greater and thus the period will be shorter than the mass-string pendulum.

 

[OlderDan]

For small oscillations, the angular frequency of a pendulum is:

$$\omega = \sqrt{\frac{g}{L}}$$

where g is acc. of gravity and L is the length to the center of mass. The conclusion that Galileo found was that for said small oscillations, the mass of the pendulum had no bearing to the oscillation period, only the length.

So for your question, consider what "L" is for both situations. For the mass on a string, it's usually assumed that the string has negligible mass, so the length would be 1m. However, for the meter stick, the mass is not concentrated at the end. In fact, it's concentrated at the center of mass, which, if the meter stick has equal mass density throughout, should be in the middle, i.e. 0.5m. Since the meter stick has a smaller "L," then the angular frequency will be greater and thus the period will be shorter than the mass-string pendulum.

This is not correct. A physical pendulum has distributed mass, and the simple pendulum equation does not apply. By your reasoning, if the meter stick were mounted and free to rotate at a point near its center the frequncy of oscillation would be higher than when mounted at the end, and would be the same as for a very short simple pendulum. In fact, the frequency would be much smaller when the stick is mounted near the center as compared to the end. In the limiting case of mounting at the center, the frequency of oscillation goes to zero (infinite period).

 

[Jelfish]

I'll admit that my prescription of that equation for the meter stick may be misguided, but in my meager defense, the center of mass that I meant to refer to would depend on where the mounting point was located. That is, the center of mass would be defined from the portion between the mounting point and the end. If it's not mounted at one end, then there would be effectively two pendulums that are linked by their angle (ok.. this is starting to sound like nonsense).

Anyway, I was going more toward the concept of how a simple pendulum acts as an upper limit case of mass distribution for pendulums of a specific length and the idea that if another simple pendulum were used to model the motion of the meter stick mounted at the end, it would make sense that it would be shorter.

Please let me know if I'm missing something intuitively.

 

[OlderDan]

Anyway, I was going more toward the concept of how a simple pendulum acts as an upper limit case of mass distribution for pendulums of a specific length and the idea that if another simple pendulum were used to model the motion of the meter stick mounted at the end, it would make sense that it would be shorter.

Please let me know if I'm missing something intuitively.

Go back and look at the general formula for the period of small oscillations posted by dextercioby in #2. That equation comes from equating the restoring torque due to the gravitational force acting at the center of mass of a rigid body, free to rotate about some axis, to the product of moment of inertia and angular acceleration about that axis (Newton's second law for rotation). You cannot separate the masses of the different parts of the rigid body, or exclude part of the mass in the calculation. The problem is one of rotation of the entire body. If the point of suspension is at the center of mass, there can be no restoring torque and the period becomes infinite.

It is true that the simple pendulum is a limiting case of the general physical pendulum, but if you want to find the simple pendulum that has the same period as some physical pendulum the length of the simple pendulum must be

$$\l_{eq} = \frac{I}{md}$$

The equality is guaranteed if the pendulum is a point mass at distance $$\l = d$$ from the point of suspension. For a pendulum that is a stick or rod of length L, the moment of inertia about a point at distance d from the center of mass is, by the parallel axis theorem,

$$I = \frac{mL^2}{12} + md^2$$

so the equivalent length of a simple pendulum would be

$$\l_{eq} = \left[\frac{L}{12d}\right] L + d$$

I don't see anything intuitive about this result. The length of the equivalent simple pendulum clearly does not correspond to the position of the center of mass (d) and is influenced by the entire length no matter where you support the stick. As a function of d, the equivalent length has a minimum value at $$d = \frac{L}{\sqrt{12}}$$, which is

$$\l_{eqMin} = \frac{L}{\sqrt{3}}$$

This position for the support minimizes the period of oscillation of the stick. You can calculate that minimum period using the equivalent length in the period equation for a simple pendulum.

 

[Jelfish]

I understand now. Thank you for the explanation.

 

[Ryker]

Sorry to ressurect such an old thread, but what happens if you keep the mass of the bob the same, but instead assume that the mass of the string isn't really negligible? I guess the pendulum then goes from a simple one to a physical one, but what happens to its period? I guess the added mass cancels out in the equation and the moment of inertia decreases. That by itself would make the period of oscillation shorter, but "d" in the equation also decreases. Does that then cancel out the effect or is the time period still shorter? My guess is the latter, but I'm having a bit of trouble wrapping my head around this so as to apply it for any case of string and a ball (that is, not having to know the exact center of mass of the new system and the precise equation for its moment of inertia).