Month: October 2014

059 Coupled Inverted Pendulum

The oscillation of one pendulum sets the horizontal tube vibrating at the same frequency. The tube is thus shaking the other pendulum at the same frequency as the first ball. In other words, all the other pendulum experience a periodic driving force at the frequency of the first ball.

Resonance occurs only for the pendulum with the same length since its resonant frequency matches the frequency of the driving force. Which is why only one ball goes into large amplitude oscillation.

058 William Tell Overture and Newton’s Pendulum

If ONE BALL rams into the stack, why must ONE BALL come out the other end at the same speed? Why can’t TWO BALLS come out at HALF the speed, for example? (This outcome also conserves the total momentum!) And how about all FIVE BALLS swinging away at 1/5 the speed?

True enough. There are infinite number of outcomes that fulfil the Principle of Conservation of Momentum! However, since these collisions are elastic, the total kinetic energy should be unchanged after the collision as well. With this additional consideration, we realize there is only one possible outcome. The same number of balls always come out the other end at the same end, conserving both total momentum and keeping total kinetic energy unchanged.

Now are you ready for Psy?

057 When 50 grams lift 100 grams (Static Equilibrium)

When the weight was hung onto a vertical string, the tension in the string was only equal to the weight.

However, when the weight was hung on a horizontal string, the horizontal string cannot remain horizontal. This is because a horizontal string cannot provide any upward component of tension to balance the downward pull of the weight.


At equilibrium, the string makes an angle θ with the vertical.

Since the vertical forces at point X should balance, 2Tcosθ = mg.

Do you see that the tension in the string is actually larger than the weight?

At point Y, we know that T = Mg. so we can work out that 2Mgcosθ = mg, which leads to the result that cosθ = m/(2M). In this video where m/M is 0.5, θ works out to be 75°.

055 When Newton Meets Einstein

The picture of Newton was in green only, while the picture of Einstein was in red color only. Green and red light from the LCD projector are polarised perpendicularly with each other. As polarizer was rotated, it cut off either the green or the red so that only either Newton or Einstein comes into view.

Some 3D movies are made this way. The image meant for the left and right eyes are projected using two mutually perpendicular polarised light. The viewers are then given glasses which are fited with polarizers.


053 Polarisation (Three Polarizers)


After polarizer 1, the light polarized at 0° with amplitude E0 and intensity I0.

After polarizer 2, the light becomes polarized at 45°, and its amplitude is reduced to E0cos45° and intensity 0.5I0

After polarizer 3, the light becomes polarized at 90°, and its amplitude is reduced to (E0cos45°) cos45°=0.5E0 and intensity 0.25I0

By inserting polarizer 2, no two consecutive polarizers are perpendicular, and the light is never completely wiped out at any stage.

052 Polarisation (Two Polarisers)

EM wave

A light wave has E-field oscillating at right angle to its direction of propagation. A polarizer allows only the component of E-field in the polarizer’s orientation to pass through.


So the first polarizer polarizes the unpolarized light. Because an unpolarized light has light in all orientations, its intensity always drops to half after passing through the first polarizer, regardless of how the polarizer is oriented.

If the second polarizer is oriented at the same angle as the first, then 100% of the polarized light is passed through. If the second polarizer is perpendicular to the first, then 0% of the polarized light is passed through. If the second polarizer is misaligned with the first by an angle θ, the amplitude of the polarized light will drop from E0 to E0cosθ, and the intensity will drop from I0 to I0cos2θ.