a. p. matthews

The difference between real, user and sys time is something I’ve long had a vague understanding of, but I don’t like things to be vague when I can help it.  So I came up with some experiments to try and produce results based on my understanding:

this program just counts to a billion:

int main(int argc, char *argv[]) {  int i;  for ( i = 0; i < 1000000000; i ++ );  return 0; }

result:  lots of user time, little sys time:

real 0m2.736s user 0m2.733s sys 0m0.001s

this program calls stat, a system call, 10 million times:

#include <sys/stat.h> int main(int argc, char *argv[]) {  int i;  struct stat buf;  for ( i = 0; i < 10000000; i ++ )    stat("/dev/null", &buf);  return 0; }

result: lots of sys time, little user time:

real 0m5.591s user 0m0.287s sys 0m5.297s

what if we just tell it to sleep a while?

int main(int argc, char *argv[]) {  sleep(5);  return 0; }

result:  little user time, little sys time, but still takes real time:

real 0m5.001s user 0m0.000s sys 0m0.001s

“Hello I would like a Double Bel please.”

“That'll be 0.20 millipascals.”

“Sorry I meant Double Del.”

The other day I surrendered to an impulse buy at the grocery store: some Swiss Miss hot cocoa mix. As I read the instructions after returning home, I realized I was ill-equipped for this impulsive purchase as I had no measuring cup to measure the the 8oz of water in Swiss Miss’s requirements. Sure, I could just add water to taste but I’m like an engineer or something, right?

Problem

Though I have no measuring cup, I do have other non-measuring cups such as this one:

It’s a cylinder with a couple things to account for: 1. The diameter increases from bottom to top 2. The base is solid

Using a measuring device I do have (a ruler) I can measure the top and bottom diameters, the height of the glass, and the height of the base. I also know I need 8 ounces of water, which is 227 grams of water. My only unknown is how high to fill the glass. Thus the problem takes shape:

Solution

The height of the water will depend on its volume. Volume (V) is inversely proportional to density (ρ) by mass (m), and the density of water (ρ_w ) is 1 gram per cubic centimeter. Thus:

The volume of a cylinder from y_0 to y_1 is:

A(y) is the area of the circular cross-section of the cylinder at y. The area is:

r(y) is the radius of the cylinder at y. Since the radius increases linearly we can use the equation of a line for which the slope would be

So in slope-intercept form, r(y) would be:

Plugging everything back in we end up with an equation relating all our quantities:

This integral can be solved by hand but the problem is we end up with a cubic equation that needs to be solved for y_1. And there’s a constant term, so it’s going to get pretty ugly having to use synthetic division. What’s a lazy engineer to do? Use the computer to solve it. In my case I use sagemath

m = 227 pw = 1 y0 = 1.75 h = 14.5 d0 = 5.6 d1 = 6.8 var(’y,y1’) solve(m/pw == integral(pi*((d1-d0)/(2*h)*y + d0/2)^2, y, y0, y1), y1)

sagemath gives me back three answers for y1, which makes sense as it’s a solution to a cubic. Only one of them is a real number and since I’m not trying to fill the glass with imaginary water, I know it’s the answer I’m looking for: 9.45cm.

Conclusion

On a local Facebook group I’m a member of, a dad asked for help with this question from his kid’s homework:

Keiko spent the day bird watching and counted 34 more birds in the morning than in the afternoon.  If she counted a total of 76 birds, how many birds did she count in the afternoon?

Since I am a terrible person I offered him this solution:

Let n ∈ ℕ be the number of birds counted in the afternoon. Let m ∈ ℕ be the number of birds counted in the morning.

Thanks, I don’t think about my ocremix contributions much anymore, but it’s nice when I’m reminded in a positive way.

I read an article on algorithmic composition from Music and Mathematics which is available here:

Universal Prediction Applied to Stylistic Music Generation

It covers a simple yet effective machine learning algorithm based on data compression that can be applied to algorithmic composition.  They even included some pseudo-code that I was able to translate easily into Python.

A non-incremental variation of this code is here which uses the example corpus from the article "abababcabdabcdabce" and creates strings of length 10 from it based on its probabilities.

Rather than jump right in to generating music, I used all The Dude's lines from The Big Lebowski as a corpus.  For your enjoyment (or bemusement) here's 10 algorithmically generated lines of dialog of The Dude:

He yelled money all over town including my rug in cash?

Uh and they're gonna kill her!

We're not that yoga?

Is and her Walter what's in the parlance two of police just hey Walter Walter what the fuck are you man!?

Just tell her fucking synagogue.

This is the driver already spent the achievers.

The fucking nihilist ah that some creedence already spent thing?

Peed on my rug in the briefcase.

Amendment fucked it home the friend is this him down a crip...you fuck sympathy!

Wait let me explain something about that shit.

Straight line AB connects two point sources that are 5 meters apart, emit 300 Hz sound waves of the same amplitude, and emit exactly out of phase. What is the shortest distance between the midpoint of AB and a point on AB where the interfering waves cause maximum oscillation of the air molecules?

Since the waves are traveling in opposite directions towards each other but their sources are exactly out of phase, this situation is the same as if they were traveling in the same direction in phase. So we can use this formula relating interference to path length difference:

where fully-constructive interference occurs when φ = m(2π) for m = 0, 1, 2,...  ∆L is the path length difference and λ is the wave length, which is related to frequency and speed of sound by λ = v/f. We’re given the frequency and speed of sound is known so we can rewrite the formula as:

The first instance of fully-constructive interference occurs when φ = 0, which would require ∆L = 0. At the midpoint between AB, the path length difference ∆L is in fact zero. Therefore the shortest distance between the midpoint and where the interfering waves cause maximum oscillation is at 0 meters.

What is the second shortest distance?

Let’s say the midpoint of AB is at x = 0, where point source A is at x = −2.5 meters and point source B is at x = +2.5 meters. Then ∆L is related to x by

where x is the distance from the midpoint. The second instance of fully-constructive interference occurs when φ = 2π. Let’s solve for x:

What is the third shortest distance?

The third instance of fully-constructive interference occurs when φ = 4π. We can use the same formula as before:

A pipe 0.6 meters long and closed at one end is filled with an unknown gas. The third lowest harmonic frequency for the pipe is 750 Hz. What is the speed of sound in the unknown gas?

The formula for harmonics of a pipe with one closed end is:

where v is the speed of sound, L is the length of the pipe and n = 1,3,5,... We’re given the length of the pipe and the frequency of the 3rd lowest harmonic (n = 5). Let’s solve for v:

What is the fundamental frequency for this pipe when it is filled with the unknown gas?

Now that we know the speed of sound in the gas we can use the original formula at n = 1 to find the fundamental frequency:

You can estimate your distance from a lightning stroke by counting the seconds between the flash you see and the thunder you later hear. By what integer should you divide the number of seconds to get the distance in kilometers?

This may actually be useful during a thunderstorm. We take the speed of sound as 343 meters per second, which is 0.343 kilometers per second – roughly close to one third. So we should be able to divide by 3 to approximate.

Two sound waves with an amplitude of 12 nm and a wavelength of 35 cm travel in the same direction through a long tube, with a phase difference of π/3 rad. What is the amplitude of the net sound wave produced by their interference?

The sum of two waves traveling in the same direction of equal amplitude and frequency differing by phase φ is

We’re only interested in the amplitude of the resulting wave, so let’s just take a look at the inside of the brackets and plug in our givens

What is the wavelength of the net sound wave?

When two identical waves (aside from their phase) interfere the resultant wave is only louder or quieter; the frequency and thus the wavelength remain the same. Another way to look at it is from the derivation: The angular wave number k is the same in the resultant wave as the original waves and k is related to wavelength by λ = 2π/k. No change of k means no change of wavelength λ.

Therefore the wavelength of the net sound wave is 35 cm.

If instead, the sound waves travel through the tube in the opposite directions, what is the amplitude of the net wave?

Two identical waves moving in opposite directions produces standing waves. The formula for a standing wave is:

As you can see in a standing wave the amplitude varies by position x. But the maximum displacement for the overall wave, i.e. the amplitude, is double the original. Therefore the amplitude of the net wave is 24 nm.

What is the wavelength of the net wave?