Month: March 2014

020 Adiabatic Compression

There are are two ways to make a gas hot. This is summed up quite neatly by the First Law of Thermodynamics:

ΔU = Q + Won

This law says that ΔU, the increase in internal energy of a system, can be brought about by

(1)    Q, supplying heat to the system, or/and

(2)    Won, doing work on the system.

a-level-physics-notes-work-done-by-an-expanding-gas1-html-2cc39decMacroscopically speaking, to do work on a gas is to compress it. Yup, work must be done against the gas pressure to force the gas into to a smaller volume.


Microscopically speaking, compressing a gas must involve a surface, let’s say a piston, charging into the gas. During this process, the piston is like a giant tennis racket smashing back all the trillions of tennis balls, i.e. the gas molecules, that happen to collide into it.


These gas molecules were sent flying back at higher speed than before. Since the average KE of the gas molecules increases, and the temperature of the gas increases.

019 What are beats?

The phenomenon of beats is a wonderful demonstration of interference of sound waves. The key to producing beats, is to have two sound waves of equal amplitude, but slightly different frequencies.

For the sake of discussion, let’s consider the superposition of a 20 Hz wave with a 21 Hz wave, illustrated in magenta and red respectively in the diagram below. Are they going to interfere constructively or destructively? Both actually. Because the two waves have slightly different periods, they actually alternate between constructive and destructive interference.


As illustrated, at t = 0.0 s, the two waves are exactly in phase, constructive interference results in a wave of twice the amplitude (illustrated in blue) and LOUD sound. The two waves then progressively drift out of phase. At t = 0.5 s, the two waves are exactly in antiphase, resulting in destructive interference, zero amplitude and SILENCE. The drifting of phase relationship continues until at t = 1.0 s, when the two waves are exactly in phase again and they are back to constructive interference and LOUD sound.

This pattern repeats once every 1.0 sec, resulting in alternating loudness and softness, called beats, at a frequency of 1.0 Hz.

With a bit of logical reasoning, we can deduce that two waves of frequency f1 and f2 will result in a beating frequency of |f1-f2|.

P.S. It’s interesting to note that when the difference in the two pitches is small, our ears hear only one pitch (the averaged frequency) with beating. When the pitch difference is large enough, our ears can distinguish that the sound is composed of two pitches (with no beating). This is something our eyes can’t do: Musicians can make out the different notes that make a chord. But a painter cannot tell if a spot of yellow is monochromatic yellow or superposition of green and red.

P.P.S. Destructive interference is how noise-canceling earphones work. An anti-phase version of the ambient noise is electronically generated to superpose with the ambient noise. Amazing!

018 Why did the shadow disappear?

It is funny that the first explanation that comes to many A-level students’ mind is diffraction grating and slit separation. The actual explanation is something much more elementary.

The changing shadow comes about because the light source was an elongated fluorescent tube. So the shadow cast depends on the orientation of the tube with respect to the comb.


In one orientation, the light source is just narrow enough to cast a thin umbra. In the other orientation, the light source is so wide it casts only the penumbra.

017 Who turned off the pressure?

Pressure comes about because of gravitational pull on the liquid. If the fluid has no weight, then the fluid molecules will not be pressing against the bottom and sides of its container, and against one another.

On Earth, we can simulate zero gravity conditions in the cup simply by dropping it. When both the water and the cup are free falling together, the water stops pressing against the cup. The water basically “hovers” in the cup and  becomes “weightless”. NO WEIGHT, NO PRESSURE. This explains why the water stops leaking out through the hole once the cup is dropped.

Likewise, hydrostatic pressure does not exist in outer space, or at the International Space Station (ISS). In fact, water hovers in the air as water balls in zero-g conditions. (Click here for proof)

016 What causes the fluctuations in loudness?

Does adding two sounds together result in a louder sound? Not always. It depends on the phase difference between the two waves.


In this video, the two speakers are emitting sound waves of the same frequency. So between the two speakers, we have two waves travelling across different distances before they arrive and superpose at each location.


At locations where the waves arrive in-phase (pressure antinodes), they interfere constructively, resulting in an even louder sound.


At locations where the waves arrive completely out-of-phase (pressure nodes), they interfere destructively, resulting in softer sound or even silence.


In this video, one could tell that the distance between two nodes (or two antinodes) is about 20 cm. This makes sense because the wavelength of the sound wave used was 42.5 cm (800 Hz @ 340 m s-1). The distance between two nodes (or two antinodes) in a standing wave is λ/2.


1. One may wonder why we did not find more nodes and antinodes between the two speakers. Well, remember the intensity of a point source is inversely proportional to square of distance from the source. So the amplitudes of the two sound waves were comparable only in the middle section. Outside the middle section, the sound from the nearer speaker overwhelms the sound from the further speaker (so the resulting amplitude from constructive and destructive interference are not very different).

2. The more observant may realize the positions of the nodes and antinodes seem to be shifting as time progresses. This is true. The two speakers were not completely coherent and their phase relationship did drift a bit. But their phase difference was changing slowly enough for an observable interference pattern.

015 Why can’t ohmmeters work together?


For ease of discussion, let’s call the Digital Multimeter (DMM) on the left DMM A, and the one on the right DMM B.

To measure resistance, DMM A actually drives a constant current I0 through its terminals. It then measures the potential difference V0 across its two terminals. Based on V0, the DMM reports the resistance R0.


What happens when DMM B joins in the fray? Since DMM B will also be driving a constant current I0 through its terminals, the total current will be 2I0,  resulting in a pd of V0=2I0R0 across R0. But both DMMs have no idea of each other’s presence and assumes a current of I0 through R0. They thus (wrongly) reports a resistance of 2V0/I0 = 2R0.


When DMM B’s range was increased to 2 MΩ, it expects a larger resistance, so it drives a smaller current of 0.1I0 instead. This results in a total current of 1.1I0 through the resistor, resulting in a pd of V0=1.1I0R0. DMM A, which assumes a current of 1I0, reports a (less inaccurate) resistance of 1.1R0. But DMM B, which assumes a current of 0.1 I0, reports (an even more inaccurate) resistance of 11R0.

The moral of the story is, when using the DMM as an ohmmeter, it must be the only EMF source in the circuit. If not, the readings will all be screwed up.

014 What is the air pressure in the bottle?

It’s all about the pressure of the air in the bottle, Pair.


Clearly, Pair + h1ρg > Patm, if not water will not be squirting out of the bottom hole.

Also, Pair + h2ρg < Patm, if not air will not be “sucked” through the upper holes.

Putting the two inequalities together, we get

Patm – h1ρg < Pair < Patm – h2ρg.

Basically, as water drains from the bottom hole, Pair drops due to expansion of the air pocket, drawing air into the bottle through the upper holes. This top up of air molecules to the air space increases Pair, ensuring that water continues to drain from the bottom hole.

Some interesting observations

0:42 – The water squirts out at a much faster rate when the bottle is uncapped. This shows that Pair is less than atmospheric when the bottle is capped.

0:54 – The water squirts out in spurts, suggesting that Pair fluctuates between the maximum and minimum values.

2:00 – At the instant when the water level dropped below upper holes, Pair increases to Patm. This results in a sudden increase in the speed of the water stream. It also became a steady stream instead of spurts.

2:26 – The stream stops before the water level reaches the bottom hole because of surface tension of the water.

The full-length unedited version is also made available if you want to do some analysis yourself.

013 Can you solve the Resistor’s Cube?

Resistance across the cube


Let’s imagine a current I passing across the cube. By symmetry, the current will split equally at each junction. We can thus tell that the current at each of the blue, green and red resistors will be I/3, I/6 and I/3 respectively.

This means that the potential difference across the cube will be

V = I/3 *R + I/6 *R + I/3 R = 5/6 IR.

Since resistance is the ratio of V to I, Reff = 5/6 R 

Resistance across a face


If the two red resistors were not there, then the effective resistance between points a and b is simply

Rab     = ( 2R || 2R ) || [ R + (2R||2R) + R]

= R || 3R

= ¾ R

Then we notice that points c and d are at the same potential. This means even if the two red resistors were there, no current will pass through them anyway. This means the two red resistors can be removed without affecting the circuit in any way.

Resistance across an edge


By symmetry, we can tell that points c and d are at the same potential, so these two points can be shorted together without affecting the circuit in any way. Likewise, points e and f can be shorted together since they are at the same potential.


The circuit can now be redrawn as shown above. The effective resistance between point cd and ef can be simplified to

Rcd,ef = [(R||R) + R + (R||R)] || R || R

= 2R || R || R

= 2/5 R

The effective resistance across an edge will thus be

Rab     = [(R||R) + 2/5 R + (R||R)] || R

= 5/7 R || R

= 7/12 R


012 Who let the water out?

The on/off mechanism of this “fountain” is effected through the pressure of the air space in the bottle, Pair.

When the bottle is not sealed


As the water drains through the holes, air from outside enters to “top up” the air space so Pair is always Patm. Since Pair + hρg > Patm, water squirts out continuously.

When the bottle is sealed


As the water drains, the air space expands and without any “top up” of the air molecules, Pair must drop. When Pair is low enough such that Pair + hρg = Patm, water stops squirting.

By how much must the air space expand? Well Patm, is about 760 mmHg, or about 1000 cmH20. If Boyle’s Law is roughly valid, then the air space only needs to expand by 1% for Pair to decrease by 1%. This to enough to make Pair 10 cmH2O below Patm, enough to “suck” and hold a water column of 10 cm in the bottle.

P.S. Actually, the required expansion is even less because surface tension also helps to keep the water in the bottle.

011 Why is the last bulb so slow?

The real mystery in this video, is why it takes longer to light up the fifth bulb if the other four were already lit up, compared to if all five were lit together.

The explanation has to do with the fact that resistance of a filament increases with temperature. So a bulb that is already lit up has a larger resistance than a cold bulb.


In this video, when the 5 bulbs are all lit up from cold together, because they grow hot together at the same rate, their resistances are the same at any one time. So the potential difference across each bulb is always 46 V (230 V /5 bulbs).


When 1 bulb is lit up from cold, with all the other 4 already hot, the initial potential difference across the cold bulb is much less than 46 V. This is because the other 4 hot filaments, with their larger resistances, has more than 46 V across each of them, as the potential divider principle dictates.

The pd across the 5th bulb will however eventually grow to become 46 V, when the bulb has heated up and its resistance becomes the same as all the other bulbs.

Compared to a bulb with 46 V across right from the beginning, a bulb with pd increasing slowly towards 46 V takes a longer time to light up.