Cats can scratch your eyes out, but can they meow your ears out?

Recently, I came across a post looking at the question "If all the cats in the world "meowed" at the same time, how loud would the sound be?". The video is below.

Because I was proctoring tests, I had some time, so I decided to go a little further. I started with a basic assumption that cats like to be about 1 meter away from other cats. Starting here, I came across some neat mathematics.

How big?

Let us assume that the approximately 600 million cats in the world want to be 1 meter apart. If we set them up on circles with radii of 1 meter, 2 meters, 3 meters, and so on, we can put 6 cats on the first circle, 12 on the second, 18 on the third, etc. How many circles do we need?

$$6 + 12 + 18 + .............n = 6 \times (1 + 2 + 3 + .......m) = 600,000,000$$

and using the Gaussian sum formula for the first m integers (the legend about how Gauss derived this can be found here)

$$1 + 2 + 3 + ...... + m = \frac{{m\left( {m + 1} \right)}}{2} = {10^8}$$
which gets around 14,000 circles.

How loud?

Obviously, a cat further away will sound quieter, but there are more of them further away. How do we calculate that balance? As the video illustrates, we can add sound intensity levels, not decibel levels. Let us use the sound intensity level for a cat meow from 1 meter away as $${{I_{{\rm{cat}}}}}$$. Remember that sound intensity diminishes as the inverse of the distance squared. So the total sound intensity of the 6 cats in the first circle will be $$6 \times \frac{{{I_{{\rm{cat}}}}}}{{{1^2}}}$$, the intensity of the 12 cats in the second circle will be $$12 \times \frac{{{I_{{\rm{cat}}}}}}{{{2^2}}}$$. The total intensity for all 14,000 circles will be
$${I_{total}} = 6 \times \frac{{{I_{{\rm{cat}}}}}}{{{1^2}}} + 12 \times \frac{{{I_{{\rm{cat}}}}}}{{{2^2}}} + 18 \times \frac{{{I_{{\rm{cat}}}}}}{{{3^2}}} + ......14,000 \times 6 \times \frac{{{I_{{\rm{cat}}}}}}{{14,{{000}^2}}}$$.
A rearrangement gives
$${I_{total}} = 6 \times 1 \times \frac{{{I_{{\rm{cat}}}}}}{{{1^2}}} + 6 \times 2 \times \frac{{{I_{{\rm{cat}}}}}}{{{2^2}}} + 6 \times 3 \times \frac{{{I_{{\rm{cat}}}}}}{{{3^2}}} + ......6 \times 14,000 \times \frac{{{I_{{\rm{cat}}}}}}{{14,{{000}^2}}}$$.
Which gives
$${I_{total}} = 6 \times {I_{{\rm{cat}}}} \times \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ....... + \frac{1}{{14,000}}} \right)$$.
Notice the harmonic series in the parentheses. Using the definition of the natural logarithm,
$$\int_1^n {\frac{1}{x}} dx \equiv \ln \left( n \right)$$  ¹,
we can approximate the sum of this harmonic series as
$${I_{total}} = 6 \times {I_{{\rm{cat}}}} \times \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ....... + \frac{1}{{14,000}}} \right) \approx 6 \times {I_{{\rm{cat}}}} \times \ln \left( {14,000} \right) = 6 \times \ln \left( {14,000} \right) \times {I_{{\rm{cat}}}}$$
But what we hear is best measured not by the intensity level, but by the decibel level. How do we go from one the other?

Most people are familiar with the decibel scale. The decibel level of a sound with an intensity of I watts per square meter is
$${\rm{dB = 10}} \times {\rm{ }}\log \left( {\frac{I}{{{I_0}}}} \right)$$ .
Using this conversion, we get
$\begin{array}{l} {\rm{dB = 10}} \times {\rm{log}}\left( {\frac{{6 \times \ln \left( {14,000} \right) \times {I_{{\rm{cat}}}}}}{{{I_0}}}} \right) = 10 \times \left( {\log \left( {6 \times \ln \left( {14,000} \right)} \right) + \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)} \right) = \\ 10 \times 1.8 + 10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right) = 18 + 10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right) \end{array}$
.
Recognize that $10 \times \log \left( {\frac{{{I_{{\rm{cat}}}}}}{{{I_0}}}} \right)$ is just the decibel level of one cat from 1 meter away which the video says is 45 dB. All the cats in the above arrangement will give a decibel reading of (18+45) dB=63 dB.
Remember that an increase of 3 dB is a doubling of loudness, 18 dB is 6 doublings, so all the cats in the world meowing at once will sound like $${2^6} = 64$$ cats 1 meter away.

I will leave it as an exercise for the reader to show that if you reduce the distance between cats to 0.5 meters, the dB level will 69 dB and will sound like 256 cats 1 meter away.

¹ I realize that $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{{14000}} \approx \ln \left( {14000} \right) + \frac{1}{2}$$ is a better approximation to the sum of the harmonic series (look at the geometry of the area under 1/x and the areas from the harmonic series), but that extra term of 0.5 doesn't change the final result appreciably.

18 October 2015

Some of you are forced to use an iPad at your school. While it has many "whiz-bang" features that impress administrators and non-science people, you realize that the iPad is very limited. You would like your students to have a device that does what you want to do, not just what Apple thinks you should be able to do.

One of the basic concepts taught in most physics classes is the linearization of data. This concept usually comes up when looking at data gathered from an acceleration experiment, be it the distance traveled on a ramp versus time or the speed at the bottom of a ramp vs distance on that ramp. Graphing that data raw gives a curve. Just looking at the curve gives no indication whether is is a a quadratic (x^2). cubic (x^3). exponential (n^x), or other function. It is only by looking at the data many different ways that the true relationship shows.

The standard approach to this task is with a spreadsheet (I always recommend Gnumeric which is free and cross-platform). But full-featured spreadsheets on the iPad cost a lot of money and free options are limited. So, for years, I used one free app to graph data and another free app to manipulate that data. While they got the job done, the procedure was cumbersome for students.

Recently, the people at Desmos added some features that allow us to do everything in that one free app. Let me sketch what I show my students to do.

While I am concentrating on using Desmos on the ipad, all instructions are applicable for the web-based application.

After students install the Desmos Graphing Calculator app and start it, they have the option of creating an account. I don't require it, so each student makes that decision. They then add the data via the Table option as seen below. Have students experiment with two-finger squeezing and stretching to better see the structure of the data.

Step 2 manipulate the data

Now come techniques specific to Desmos. You now need to generate a new set of points based on your chosen manipulation. In this example, we want to square the independent variable but keep the dependent variable unchanged.

Start a new box on the data side. Type in what you see below (if you just type "x1". Desmos is smart enough to make the "1" a subscript). As soon as you complete the manipulation (seen below), you see the new points on the graph to the right. With some two-finger stretching and spreading on the graph, you can see the points clearly.

Now you need to bring those points from the graph to tabular form. Click on the octagram just above the data and you will see something similar to below.

Step  3 display the new points

Click on the Table icon and you will see the manipulated data.

Step  4 get the line equation

To get the linear regression equation for the above data, start a new box on the data side. Type the equation as you see below (the tilde "~" is important). Now you have the needed equation of the line of best fit.

PS Desmos can also solve quadratic equations

Another nice feature is that Desmos can solve the typical equations encountered in motion analysis. Below is the equation to determine the time that an object is 25 meters above the ground when is thrown upward from a 15 meter platform at 20 m/s. Notice that the equation does not have to be put into the standard form as per most other web-based equation solvers.

21 May 2014

Hints on "Justify and Explain" questions on the AP 1 and AP 2 tests

Below is a conversation on the AP Community website. I reprint it here because I think it needs wider exposure. The response is from a long-time AP Reader who is held in high regard by many of the other Readers.

 Justify and Explain questions... I need advice, or resources, that will help students answer ‘justify’ or ‘explain’ questions accurately and concisely (our school’s English department has tried to help, but they don’t seem to grasp the problem), and how to assess their answers accurately and efficiently. Next year, I’ll teach five periods of 40 students  (four of P1 and one of C (M,E+M)), which has always created lots of grading… but now, with the advent of the new exam, there’ll be more, especially of the ‘justify’ and ‘explain’ type. What do you do to improve your student’s answers and your assessment of them, while saving time?

Physics Teachers being Naughty

One of the nice things about being a member of the Western New York Physics Teachers Alliance is the email list. Here we can share ideas with colleagues  we may see only once every month or two at our regular meetings. The following is an example of one such exchange (names have been deleted to protect the guilty).

I've got a PVC cannon that can fire ping-pong balls/dog toys, paper tubes/rockets, and dry erase markers.  It's electrically triggered, but I'm still trying to figure out what the max pressure it can take and still electrically fire.  Individual components are good up to 90 psi, but around 70, the sprinkler doesn't actuate.  40 psi will send a dog toy at ludicrous speeds.  I also went out and got an air compressor at Harbor freight, so we can basically fire the thing every minute or two (depending).  Air compressor is loud as dickens, tho.  Actual muzzle velocity calculations/high speed forthcoming.

It can fire dog toys; can it fire toy-dogs (not the plushy kind)?

If you can get them to fit in the barrel.  And not wiggle around too much.

We could get a Doppler-shift demo out it, too.

"Yip-yip-yip-yip-yap-yap-yop-yop-yop-yop"

17 February 2014

5-g's? Let's not Exaggerate.

Watching the Olympics, I have seen a commercial that annoys me. It features a figure skater comparing what she goes through to other occupations. One is that she "experiences more g-forces than a fighter pilot". Let's examine that claim.

The math

The forces on the various parts of a spinning skater are due to rotation. However, we will look at the acceleration since that what g-forces really are. For a 5g-force, the relationship between rotation speed (RPM) and distance from rotation center (r) is

A graph might make this equation a little more meaningful.

So, we can see that a 5-g force can be obtained for a hand rotating 1 meter from the center at a little more that 60 rpm (or one rotation each second). But the head? Even with a skater's head held straight out at around 20 cm, 5-g's would only be experienced at 150 rpm (over 2 rotations per second). And if we look at a 9-g force which most fighter pilots experience occasionally (and for a lot more than a second), it would take 200 rpm. While the supposed world record for a figure skater spin is around 300 rpm, that skater held her head upright, so her nose might feel a 9-g force, her head would feel a much lower force.

Conclusion

What can we conclude? While it is possible that a figure skater can experience over 5-g's of force on some parts of her body, when was the last time you saw a skater do a rapid spin in competition? So, the ad is true, but very misleading.

Banning the C-words (the physics ones)

It has a name and even an equation, so it must be real.

One of the challenges when dealing with circular motion in a typical physics class is getting students to realize that the centripetal force is the result of other forces. One trick that I have used in my class the past few years is to use the phrase "the force needed for circular motion". By never using the term "centripetal force", I try to emphasize that we need other, real forces to make something move in a circle. This approach seems to help when looking at classic examples, like the roller coaster (see a previous post here). This approach also helps because I don't have to get into the discussion of the difference between centripetal and centrifugal. Unfortunately, for typical high school students, once you give something a name, it becomes real.

So, do like I do, and ban the C-words in your class.

24 December 2013

An extra 8 seconds on Jan 4

21 Dec is the shortest day of the year and 4 Jan is the longest.

We all know that the day of the winter solstice is considered by many the shortest day of the year, but did you know that we can also consider the day of perihelion (4 Jan) as the longest day? It all comes down to how you define the length of a day.

How long is a day?

Ask someone how long a day is and you will probably get the answer, "24 hours". However, ask someone at the US Naval Observatory and you will get "1420x106x60x60x24 vibrations of a hydrogen maser". Ask a farmer and you will get "From sunup to sunset". We have any different definitions. For this post, I am going to use the synodic definition, the time it takes the Sun to go from due south (local solar noon) to the next time it is due south.

It starts with Kepler.

Thanks to Johannes Kepler, we know that the Earth orbits the Sun in an eliptical orbit, and that the Earth is moving fastest when it is closest to the Sun (perihelion) and slowest when it when it is farthest (aphelion).

This year, perihelion occurs on 4 Jan, which will be the longest synodic day of the year.

Let's figure it out.

Since on the above diagram, the Earth is also spinning counter-clockwise, the diagram below shows (with the added lines) one complete rotation of the Earth for both the day of aphelion (3-4) and the day of perihelion (1-2). Note that the diagrams are exaggerated for effect; the Earth does not really move this far in it orbit in one day.

Notice that it takes a little more than one rotation of the Earth to have the Sun directly due south again. Also notice that the Earth has to rotate a little more at perihelion than at aphelion to be facing the Sun 1 synodic day later.  So a synodic day is longest at perihelion and shortest at aphelion.

But how much longer is it?

While the above argument is enough to show that 4 Jan 2014 is the longest (synodic) day of the year, it does not say by how much.  Is it minutes or milliseconds? To answer this, we will need some numbers and the more precise form of Kepler's Law used above.  Kepler figured out that orbiting bodies sweep out equal areas in equal time. And by looking at the diagram above, you can see that the angle swept out by the Earth in one sidereal day is equal to the extra angle the Earth has to rotate to get the Sun due south.

Since the area of a circular sector is
this gives the relationship between for angles swept in equal time.

The Earth-Sun distance is 152,098,232 kilometres (or 1.01671388 AU) at aphelion and is 147,098,290 kilometres (or 0.98329134 AU) at perihelion.  This gives

Microseconds or Hours?

So, the Earth rotates around the Sun just a little bit less on 5 July than on 22 December. Since the Earth rotates about 1 degree (actually 360/364 degrees), it rotates about 0.033 degrees less on 5 July for 1 synodic day than on 4 January. How long does it take the Earth to rotate that angle? Since it takes 24 hours to rotate 360 degrees, it takes about 8 seconds to rotate that 0.033 degrees. So the synodic day on 4 January is about 8 seconds longer than the synodic day on 5 July.

Classroom use.

Unless you are teaching summer school, this is not much use in the classroom.  However, if you flip the seasons, you can show that the day of perihelion is the longest day of the year (and it is close the the "shortest" day of the year, 22 Dec). Since the first day back from the Christmas break is close to 4-5 Jan, I use this for a gradual welcome-back class.  In a regular physics class, we just do the qualitative approach with students playing the parts of the Sun and the Earth.

The discussion starts with what causes the seasons.  It helps to have a globe handy at this point. We then discuss Kepler's laws. I get one student to play the part of the Sun and one the Earth.  Depending on time, I might have the class discuss how the Earth should rotate (the Sun rises in the east). If not, I will just get the Earth rotating in the correct way.  The class decides when the Earth has rotated one complete rotation relative to the room and then notices that it is the Earth (the student) is not facing the Sun. That 5 minute activity sets up the above discussion.

A related post (http://canisiusphysics.blogspot.com/2013/03/if-spring-starts-tuesday-why-does.html)