harmonic-oscillator
This image shows the forces acting on the bob of a simple pendulum:
This image shows the forces acting on the bob of a conical pendulum:
In the first one, $mg \cos \theta = T$, and $mg \sin \theta$ is the restoring force. In the second one, $F \cos \theta = mg$, and $F \sin \theta$ is the restoring force. Why does that difference exist?
Gravity exerts a constant force downwards. If you apply a second constant force sideways you are in effect rotating the gravitational force:
where the angle $\theta$ is given by:
$$ \tan\theta = \frac{F_\text{ext}}{F_g} $$
The modulus of the net force is given by the usual Pythagoras rule:
$$ F_\text{net}^2 = F_g^2 + F_\text{ext}^2 $$
So if you imagine mentally rotating the whole experiment an angle $\theta$ anticlockwise you would have the pendulum handing vertically downwards in an effective gravitational field of $F_\text{net}$. This should make it obvious how the period changes.
My question is, why does the length of a pendulum cause different natural frequencies of the pendulums?
An oscillator undergoes fee vibrations when no external force acts (i.e. undamped e.g. no air resistance); when a system is displaced and left to oscillate, the system oscillates at a frequency known as the natural frequency. For the case of a pendulum with small oscillations, its period is described by the following formula: $$T=2\pi \sqrt{\frac{L}{g}}$$ where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity $=9.81ms^{-1}$ (see the image below). This arises from solutions to the equation of simple harmonic motion for a pendulum (this requires knowledge of calculus), and there are many derivations online, for example here and here. However, the relationship makes sense intuitively - imagine short and long pendulums freely oscillating next to each other; the long pendulum will oscillate slowly, taking longer to move through some angle $\theta$. Likewise, the dependence on the strength of the gravitational field $g$ is intuitive - as $g$ increases the restoring force increases (the force which returns the pendulum to the equilibrium position), and so less time is taken to return to the equilibrium position (lower period).
Barton's pendulums demonstrate a different aspect of oscillatory behaviour, resonance. When an oscillator is not left to freely oscillate (as was assumed above), but is instead driven by some periodic force at a frequency $f$ (i.e. we apply oscillations to the oscillator at a specific frequency), then the oscillator will now be forced to oscillate at frequency $f$, undergoing forced vibrations. This can be understood by considering what happens when a child on a swing is pushed - the swing itself will oscillate at some natural frequency $f_0$ when displaced and left, but when we push it it will now be forced to oscillate at $f$. At some values of $f$ (e.g. if we push it very rapidly or very slowly) the swing will only have a small amplitude of oscillation (does not oscillate with a large displacement), but at other values the swing will keep getting higher and higher. The higher and higher phenomenon is known as resonance; at resonance the driving force continually supplies energy to the oscillator so the amplitude of oscillation gets bigger and bigger, limited only by the damping forces (the air resistance and friction here). It can be shown (again, calculus based) that this pehnomenon occurs when the applied driving frequency $f$ is equal to the natural frequency of the system $f_0$ i.e. resonance occurs when $f=f_0$.
Consider the image of Barton's pendulums below:
When the pendulum bob $X$ is displaced, it oscillates at its own natural frequency $f_{X,0}$, which depends on its length. This exerts a driving force on the other pendulums via the connecting string, and so these undergo forced vibrations. As described above, these oscillations will be greatest (resonance occurs) when the natural frequency of the pendulum bobs matches $f_{driving}=f_{X,0}$. Since pendulum $C$ has the same length as $X$, its natural frequency $f_{C,0}=f_{X,0}$, and so it oscillates with maximum amplitude. The other pendulums have some other frequency and so just like when we push a swing too quickly or too slowly these oscillate with less amplitude.
Interestingly, the pendulum with the same length as the brass bob will coincidently have its natural frequency as the bob's driving frequency, could this be explained?
As should hopefully be clear from above, there is no coincidence beyond us setting two of the pendulums to have equal lengths. The bob $X$ is initially displaced, and so its own natural frequency provides the driving frequency for the other bobs. Where a pendulum has the same length as $X$, the natural frequency of its oscillation will match the driving frequency, and so it undergoes resonance.
Best Answer
In the second scenario, a centripetal force is required, perpendicular to the motion of the bob, which only the horizontal component of the tension in the string can provide. In the first scenario, the restoring force in the simple harmonic motion of the simple pendulum acts towards the mean position, which is provided by the corresponding component of weight. Hence, $mg$ acts as the restoring force. They are two very different scenarios.