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In this idealization, no energy is lost to the system; therefore, it is converted from all potential energy at the top of the arc to all kinetic energy at the bottom of the arc. This is known as Rayleigh's energy method.

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Simple harmonic oscillation is a system described by

[tex]\ddot{x}=-\omega^2 x[/tex]

where [itex]\omega[/itex] is some constant, and [itex]x[/itex] is usually a displacement variable. We can solve this equation in general, very easily. An example of this is the spring-mass oscillator.

Another example is the simple pendulum, where [itex]x=\theta[/itex] the angular displacement, and [itex]\omega^2 = g/l[/itex].

As for the physical pendulum, if we are careful, we find that it is not a true simple harmonic oscillator; instead it obeys the following equation:

[tex]\ddot{x}=-\omega^2 \sin(x).[/tex]

In the case of small displacements:

[tex]\lim_{x\rightarrow 0}\sin(x)=x[/tex]

and we find that in this case, the physical pendulum is well approximated by the simple pendulum. This is useful since the SHO equation is solved much more easily than the equation for the physical pendulum.

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arildno

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The only difference between the "simple pendulum" and the "physical pendulum" is the mass distribution; the simple pendulum consists of a massless rod ending in a bob with mass.

An example of a "physical pendulum" would be a rod of uniform density.

Note that this difference in mass distribution makes for different moments of inertiae, and therefore different periods of the pendulae.

Simple harmonic motion is not present in either case, but for small angular displacements, that approximation is justified.

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Also there is a mass dependence in the physical pendulum right and does the theoretical formula hold for small angular displacements?

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Isn't this merely a case of definition? I'll accept, however, that you know that in the literature the common usage is as you described.The only difference between the "simple pendulum" and the "physical pendulum" is the mass distribution; the simple pendulum consists of a massless rod ending in a bob with mass.

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does the period have any m dependence in the simple pendulum? No right since its a masless rod. Also under what condition does the theoretical formula hold for the simple pendulum, would it be only for larger angles and amplitudes?

Also there is a mass dependence in the physical pendulum right and does the theoretical formula hold for small angular displacements?

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pawana,

do you go to stonybrook? these questions sound reeeally familiar..

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arildno

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Of course it is just a matter of definition. I furnished the one I'm used to.Isn't this merely a case of definition? I'll accept, however, that you know that in the literature the common usage is as you described.

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And also, the differential equation is linear and very easily solved.

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OlderDan

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There is no m dependence in the motion of a simple pendulum because the restoring force is due to gravity and is proportional to the mass. This is similar to the situation for projectile motion. Since F = ma, when you have a force that is proportional to mass, the mass divides out.

does the period have any m dependence in the simple pendulum? No right since its a masless rod. Also under what condition does the theoretical formula hold for the simple pendulum, would it be only for larger angles and amplitudes?

Also there is a mass dependence in the physical pendulum right and does the theoretical formula hold for small angular displacements?

One can (and people have) engaged in debate over whether the gravitational mass in GMm/r² and the inertial mass in F = ma are the same thing, but for all the situations you are likely to encounter, they can be safely treated as being the same.

The mass dependence neatly drops out for the simple pendulum if you treat the mass as a point object that has no rotation. If the mass is distributed, then the distribution of the mass comes into play. Even a small spherical bob would have some rotation, so you could treat it as a physical pendulum and find a slight difference in the calculated period compared to the simple pendulum. However, if you look at the equation for the period of the physical pendulum it involves the ratio I/m and I is always proportional to the mass of an object. So the mass divides out for the physical pendulum as well, but the factors in I that depend on the shape of the object do not drop out

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

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