These puzzlers test your intuition into physics. See if you can solve them before reading the solutions.
Have you ever noticed that the bumpers on a billiard table are slightly too tall?
Now I'm not a billiard player, so I don't have much experience with billiard tables, but every one I've ever seen has had the bumper edge at a height h that is above the center of the billiard ball, as shown above. By this I mean that the bumpers are taller than the radius of a billiard ball, h>r. Perhaps this is to keep the ball on the table during a high speed impact with the bumper? Yeah, this makes sense -- with the bumper higher than the center of mass of the billiard ball, the collision would tend to direct the ball down onto the table top instead of launching it into the air. [Of course, if this were true then the downward force on the ball might make it "bounce" off the table's surface and still be launched into the air. What do you think?] But did you ever consider that the height h of the bumper may serve another purpose?
In fact, the bumper may be designed so that a ball caroming off a bumper does so at an angle equal to its angle of incidence! I know, you're probably thinking that this is obvious --- that the ball will always ricochet at the same angle as long as English is not involved. But no, it isn't obvious and, in fact, if the height h were not carefully chosen the angles of incidence and rebound would not be the same! Can you explain this? Can you calculate the proper height of the bumper?
I propose the following mechanism. A ball without English (no spinning about a vertical axis) directed at the table bumper at an angle of theta_i will ricochet off the bumper at an angle of theta_f. If the billiard ball "slides" on the table surface during this ricochet, then since "sliding" leads to increase friction between ball and table top the velocity of the ball parallel to the bumper after the bounce will be less than the velocity of the ball parallel to the bumper before the bounce, as shown in the above drawing. In other words, sliding friction causes the parallel velocity of the bouncing ball to decrease. Assuming the bumper provides a fairly elastic rebound, meaning the the velocity of the rebound perpendicular to the bumper is just opposite to the perpendicular velocity before the strike, then in this case, theta_f < theta_i, that is, the ball does not ricochet off the bumper at the same angle as it struck the bumper. [See the above diagram.] Therefore, I claim that the height of the bumper is determined not by the above described need to "keep" the ball on the surface by directing a bumper force downwards, but rather by the requirement to have a billiard ball with no English ricochet off the bumper at the same angle as its angle of incidence. What do you think?Puzzler: Assuming that my mechanism is valid, compute the height h necessary to accomplish this theta_f = theta_i ricochet.
When I measured the bumper height h on a neighbor's billiard table and the radius of his billiard balls r, I found a close match between the above theoretical value and the actual height of the bumper. After you compute your theoretical h, measure the h on a billiard table. How closely do they match? I would be interested in hearing about your results, as I only have measured a couple billiard tables. Thanks.
Here is the calculation of the theoretical height:
(If you are reading the Billiard Bumper Bounce Puzzler because you linked here from the The Most Marvelous Mechanical Machine Demonstration page, then click the following link to return: FPF-03 The Most Marvelous Mechanical Machine .)
Here is a yo-yo swinging off the end of an electric screwdriver. What happens when Syd hits the button and turns on the screwdriver? Will the yo-yo then begin to rotate its plane of swinging, or perhaps convert over to a more elliptical path, and thus strike the wall behind her? Or will it simply keep swinging in the same plane in front of the wall behind her, even when the electric screwdriver is spinning?
Okay, maybe you didn't like that example, or perhaps you were worried about the fact that the yo-yo was attached to the electric screwdriver by a string that can readily twist itself? Well, how about the following experiment. Below Syd is flexing a stiff thin steel rod so that it is vibrating in a vertical plane, as you can see in the two photographs if you look closely. She is standing in front of a garage door so that you can also see the shadow of the thin rod on the white garage door background as it oscillated back and forth. Notice that the steel rod is held in the chuck of an electric drill. Now what happens when Syd turns on the electric drill? Will the vibrating rod twist around and strike the garage door as it swings back and forth?
Puzzler: Does the swinging yo-yo or the vibrating rod, or both, strike the wall behind them when the electric drills are switched on?
Here is the answer and physical explanation:P.S. The mechanism behind this experiment is the same mechanism that your iPhone 4 uses to measure rotations of the device when playing games! So, when you understand the above swinging yo-yo and vibrating rod experiments, you also understand how your iPhone, Wii, and Xbox Kinetic work.
(I may demonstrate these experiments at the FPF talks.)
Ocean tides have always been somewhat of a mystery, mostly because the simplest explanation appears to predict the wrong behavior.
If ocean tides are due to the gravitational tug of the Sun or Moon on the water in the oceans, then why are there two high tides each day? It is fairly obvious that the water would be pulled towards the Sun or Moon, but why is there a similar water bulge on the opposite side of the Earth?
Since the days of Newton's Law of Universal Gravitation, we have known that the gravity force from the Sun on the Earth holds the Earth in its orbit. This gravity force is between every particle of the Sun and every particle of the Earth, including the water in the oceans.
We know from Newton's gravity law that between any two objects, say 1 and 2, having masses m_1 and m_2, the force of gravity is
where G=6.67259x10^{-11} N m^2/kg^2 is the Universal Gravitation Constant, and r is the distance between the centers of mass of the two objects. This formula says that the force of gravity between two bodies is proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance separating the centers of mass of the two bodies. This force acts along the line connecting the centers of mass and is always attractive.
Notice that Newton's Gravity Law tells us that as the distance between two objects increases, the gravity force between them decreases.
So the argument for the tides is that the side of the Earth facing the Sun (daytime side) is obviously closer to the Sun than the opposite side (nighttime side), thus the Sun's gravity force is greater on the particles of the Earth closest to the Sun than on the particles on the nighttime side. Most of the particles of the Earth are held in place since the Earth is solid. The oceans, however, are liquid and thus the water in the oceans can move under the influence of the Sun's gravity. Since the Sun's gravity is strongest on the side of the Earth closest to the Sun, the ocean water there will be pulled more strongly towards the Sun. In other words, the ocean water will be pulled up directly under the Sun resulting in a bulge in the ocean on the side of the Earth facing the Sun, that is, in a daytime high tide. The water in the ocean on the nighttime side of the Sun will flow around to the daytime side, producing a nighttime low tide. The solid Earth then spins on its axis once a day underneath the ocean bulging towards the Sun. But this explanation would predict a high tide once a day, while any mariner can tell you that ocean high tides occur roughly twice a day, not once a day.
Another explanation predicted that high tide should be on the nighttime side of the Earth. The argument is that the solid mass of the Earth is closer (i.e., the center of mass of the solid Earth) to the Sun than the water in the ocean on the side opposite to the Sun. Thus the solid Earth experiences a stronger gravity force from the Sun than the ocean water on the nighttime side. Hence the Earth is "pulled away" by the stronger gravity force of the Sun from the far side ocean, producing a bulge in the ocean on the side opposite to the Sun. Essentially, the Earth is pulled away from the ocean by the Sun, producing a high tide on the nighttime side of the Earth.
Combining these two explanations would give two high tides per day. In summary, the water in the ocean closest to the Sun is pulled towards the Sun more strongly than the solid Earth's center of mass, being farther away, producing a bulge (high tide) on the daytime side of the Earth. Likewise, the solid Earth is closer to the Sun than the far side ocean and thus the solid Earth is pulled toward the Sun with a stronger force than the far side ocean water, essentially leaving behind an ocean bulge (high tide) on the nighttime side opposite the Sun. This combined explanation thereby yields two high tides per day, as is observed in the above drawing.
Unfortunately, both of these explanations, and their combination, are wrong (or, at a minimum, they are not complete in detail which leads to a misunderstanding of the physics behind the tides)! In fact, the Nobel laureate Richard Feynman states in his The Feynman Lectures on Physics, "Both of these theories are wrong". Intriguingly, the above argument for the tides is essentially that given by Newton in his Principia. Nevertheless, can you give a more proper physics explanation of the ocean tides? [I feel that Feynman was being a little too harsh in his comment, although other physics textbooks make similar statements about these tide explanations. There is at least a partial validity to these explanations, it is just that they don't tell the entire story and thus sometimes lead to an improper interpretation of the mechanism for tides. In this Example we will give a fuller picture of the nature of tides and even compute their amplitudes.]Puzzler: Please provide a more proper proposal for the mechanism of the tides. Can you compute the actual height of the tides in the middle of the ocean away from any shoreline?
Hint 1: Recall that the radial acceleration, a_R, for circular motion is given by
where v_T is the tangential velocity and r is the radius of the circular motion.
Hint 2: Or, you might calculate the actual forces in the noninertial Earth frame of reference thereby computing the height difference between low and high tide.
Here is the explanation for why the above two arguments are not correct, as well as a calculation of the actual height of ocean tides:
What if the Moon were slowing down, which it is by the way, would it loose energy and eventually spiral in and strike the Earth?
Or maybe because it is slowing down, it will slowly drift away from the Earth?
The Apollo astronauts left a highly precise retroreflector [How does a retroreflector work, anyway? If you just put a mirror on the Moon, a light ray from the Earth would be very unlikely to reflect directly back to the Earth.] on the Moon, and by beaming lasers off this reflector it is known that the Moon is ever so slowly drifting away from the Earth (about 3.8 cm per year).
We also understand that Conservation of Momentum, both linear and angular, is the very foundation upon which all of physics is constructed. Since the Moon is moving away from the Earth, then how can this be juxtaposed with the Conservation of Angular Momentum Law? If angular momentum is conserved, then the Moon should not be drifting away from the Earth. Now the Moon is moving very, very slowly away from the Earth, so maybe the Conservation of Angular Momentum Law is only approximately valid, just like Newton's Laws of Motion? At most practical scales we observe angular momentum to be conserved, but at extreme accuracies would we find that angular momentum is only approximately conserved? This is similar to the way Newton's F=ma works for most everyday work, but is not valid when speeds approach the speed of light.
No, Conservation of Momentum is believed to be exactly true even in Einstein's Relativity and Quantum Mechanics.Puzzler: How do you justify the experimental fact that the Moon is slowly moving away from the Earth with the Conservation of Angular Momentum Law?
This PDF file discusses how Conservation of Angular Momentum can be maintained by the Earth/Moon system:
Puzzler: Given two identical cameras (or two eyes) separated by a distance D, from measurements made of distances in the two images (photographs), find the 3-space location of the object.
And the solution is:
Have you ever noticed that you can spin a paperback book around two of its principal axes smoothly without trouble, but if you attempt to rotate the book about its third axis it will tumble?
Click for full-size image Syd is attempting to catch a tumbling paperback book...Puzzler: Give both a Topological Argument as well as a Quantitative Argument as to Why Tumbling Occurs. Also explain about which axis the tumbling occurs.
Did you know that the reason this occurs is partially to blame for the 3 year delay in Gravity Probe B's experimental verification of the Lense-Thirring (frame dragging) effect from Einstein's General Relativity.
Here is a description of the Gravity Probe B experiment:
It is also the same reason why a quarterback must have talent so that he can throw a pass without the football tumbling.
Hint 1: Think about the topology of the following 1-dimensional phase diagram:
Notice that in 1 dimension, for any smooth function, f(x), then between any two minima, say at x_1 and x_2, there must be a maximum at some point x_{\star}. In other words, you cannot have two minima in 1 dimension unless they are separated by a maximum.
Hint 2: Apply the same type of topological thinking as employed above for the 1-D function to the 3 dimensions of the following phase diagram (frictionless):
...and then extend it to (with friction):
Here is the topological analysis of this problem as well as the quantitative calculations:
Perhaps this is not even possible!?And why would we even care?By watching a gyroscope, we notice that almost any movement produces small torques on the axle of the gyroscope that lead to slow precessional motions, called polhode motions, of its axis (see the photograph of a gyroscope resting on a rotating platform in the photograph below.Thus the gyroscope's axis does not continue to point in a stable direction in the inertial frame as it moves. In fact, no matter how the gyroscope is designed and manufactured, it is impossible to eliminate these polhode motions completely. A case in point is the Gravity Probe B satellite containing four of the most perfect gyroscopes ever manufactured (see the "Gravity Probe B" topic in the PDF below for a description of this experiment). These gyroscopes are nearly perfect quartz spheres, coated with a nearly uniform superconducting niobium film, suspended by an electric field, spinning in a nearly perfect vacuum, and measured by SQUID magnetometers. But even these nearly perfect gyroscopes have experienced polhode motions that have caused much more analysis and numerical modeling than was initially expected and consequently these polhode motions have delayed the publication of the experimental Lense-Thirring results testing Einstein's General Relativity theory for over three years.
Click for full-size image Here is a description of the Gravity Probe B experiment:
So, apparently it is nearly impossible to make a perfect gyroscope, at least perfect enough to provide a stable platform against which to measure any movements relative to the inertial frame for long periods of time (e.g., the Gravity Probe B's gyroscopes were used to measure General Relativity's geodetic and Lense-Thirring effects against the inertial frame for a period of 10 months).
Most long term navigational requirements, however, occur on or near the surface of the Earth, and thus typically only require 2-dimensional positional information, not 3-dimensional data. And it turns out that the inertial navigational errors in 3 dimensions stemming from system errors are unbounded and must be adjusted by external positional aids over long times, while in 2 dimensions these errors can be bounded and thus accounted for and controlled providing an unaided inertial navigational system. This bounding of the 2-dimensional errors relies primarily on the concept of a Schuler Pendulum, discovered by Maximilian Schuler in 1923.
In two dimensions, accelerometers are mounted on a platform. If this platform is horizontal, then the accelerometers measure the accelerations in the directions of travel directly, and their outputs can be integrated directly to calculate the vehicle's velocity and position. If the platform is not horizontal, however, then angle-dependent scale factor errors enter into the accelerometer calculations and drift errors enter into the orientation calculations which causes systematic errors to creep into the navigational computations. In fact, it can be shown that in 3 dimensions, the drift errors are unbounded and will cause the navigational system to seriously miscalculate its location after a relatively short time. In 2 dimensions where the 2-dimensional plane remains locally horizontal so that the accelerometers mounted on this plane measure horizontal accelerations that can be integrated without the need for scale factors, the drift errors then remain bounded and thus the navigational system will be accurate for a much longer time.
Hence we wish to maintain a stable horizontal platform on which to mount our gyroscopes and accelerometers in order to minimize the navigational errors. But this platform is aboard a ship, submarine, airplane, or land rover, and consequently is buffeted by movements in all directions, that is, the platform experiences random accelerations along any direction as the vehicle travels. If we wish to have the platform remain horizontal, then it must be stable in the face of these random accelerations. Is this possible? How is this possible?
A Little History First:
Let's begin with just a brief historical vignette, shall we? Hermann Anschutz-Kaempfe (1872-1931) founded a German company in 1905 (Anschutz & Co. GmbH, Kiel, now the Raytheon Marine GmbH, Kiel) to manufacture navigational devices based upon the gyroscope. His cousin, Maximilian Schuler (1882-1972), joined the company in 1906. One of the chief problems that plagued the company was how to establish a stable platform on a rolling, pitching, and turning vessel. Maximilian Schuler discovered the solution to this problem which he published in 1923.
But let's back up even a little more, for there is yet further connections leading all the way back to Foucault. You see, Foucault's 1851 gyroscope was employed in an 1885 patent by the Dutchman Marinus Gerardus van den Bos in a navigational device called the gyrocompass (the amazingly ingenious gyrocompass is the subject of another Example). It never really worked properly for van den Bos, however. In 1903 the German engineer Hermann Anschutz-Kaempfe constructed a working gyrocompass for which he obtained both a German and an American patent in 1908. The American Elmer Ambrose Sperry filed an American patent for his improvements on the gyrocompass, specifically the suspension and oscillation damping mechanisms, in 1911 which was later issued, with amendments, in 1918. When Sperry attempted to sell his gyrocompass to the German Navy in 1914, Hermann Anschutz-Kaempfe sued for patent infringement. Elmer Sperry's claim was that the Anchutz-Kaempfe patent was not a significant improvement on van den Bos's 1885 patent and thus was invalid. Albert Einstein (yup, the one and only) was called to testify in the court case, initially siding with Sperry. But, as was to become a recurring theme with Einstein, he reversed his opinion and sided with Anschutz-Kaempfe saying that the Sperry patent infringed upon the Anschutz-Kaempfe patent's method of damping. Anschutz-Kaempfe won the lawsuit in 1915. The Franklin Institute in Philadelphia, which houses a very large Foucault's pendulum by the way, has a very intriguing "Case File #2524" reporting details of the Sperry/Anscutz-Kaempfe conflict. This arose because Elmer A. Sperry was nominated for the prestigious John Scott Legacy Medal. Upon the nomination of Sperry for this award, Hermann Anschutz-Kaempfe sent a letter to Hugo Bilgram, the Chairman of the Franklin Institute's Committee on Sciences and the Arts which determined the winners of the Scott Medal, objecting to and critical of Sperry's nomination. The Case File includes a number of letters between Anschutz-Kaempfe, Hugo Bilgram of the Franklin Institute, and Sperry. These letters show that Sperry, by his own admission, would pinpoint the weaknesses of other inventions and then attempt to make improvements. This is precisely what he did for his suspension and oscillation damping improvements for the gyrocompass, originally patented in the U.S. by Anschutz-Kaempfe in 1906. But Anschutz-Kaempfe argued that Sperry's gyrocompass was not superior to his own and thus the Sperry patent did not constitute valid improvements.
On to the Problem:
In any case, and without attempting to navigate the muddied waters (pun intended) between Sperry and Anschutz-Kaempfe, we return to our original story concerning Max Schuler who in fact did make a significant improvement to the gyrocompass for Anschutz & Co. GmbH, Kiel. As mentioned in an earlier paragraph, the development of an accurate navigational tool necessitated the establishment of a stable platform. Difficulties arise on ships that are turning, pitching, and rolling in the turbulent seas. Each acceleration of the ship would tend to destabilize any stable platform. So, what is the engineer to do? Max Schuler solved this problem in 1923 by the expedient use of a pendulum, not just any pendulum, however, but a very specific compound pendulum.Say we start with the pendulum shown in the above Figure. If the pendulum in Part A is 1 meter in length, then it has a period of 2 seconds. What happens if we were to quickly shove the support horizontally to the right with an acceleration of a drawn as the blue arrow in Part B? If the impulse from a is large enough to produce its displacement in a time short compared to the pendulum's 2 second period, then initially the bob would attempt to stay in its place, producing a pendulum bob offset from its equilibrium position. But then gravity would take over and the bob would swing to its new equilibrium position and oscillate around it. How long would it take to swing to the new equilibrium position vertically below the new position of its support? Yup, that's right, it would take a half-second to swing into its vertical orientation, based upon its two second period and assuming a linear harmonic oscillator model for the pendulum. On the other hand, Part C illustrates the situation where the acceleration a is much smaller (green arrow), so small in fact that the displacement due to the impulse occurs in a time interval long compared with the period. The bob then does not lag begin its pivot, but it just keeps up and moves over simultaneously with the pivot.Thus we see that if the pivot's displacement is fast compared to the pendulum's period, the bob does not keep up and is offset from its vertical equilibrium position (the cable does not remain locally vertical, in other words). The bob then swings back to equilibrium in a time determined by the pendulum's period. If the pivot's displacement is slow compared to the period, the bob's movement does keep up with the pivot and the bob remains in its vertical equilibrium position during the displacement. The bob does not have to swing back to equilibrium.Now what would happen if we make the pendulum's period infinitely long? For instance, take the compound pendulum shown in Part D of the following Figure. This pendulum has an infinite period since the pivot resides at precisely the center of mass of the pendulum. Notice that if we quickly shove, or accelerate, the pivot of this pendulum as shown in Part E, the pendulum's bobs both move over simultaneously with the movement of the pivot. Thus the orientation of the pendulum remains vertical. Part F shows what happens when the pivot is slowly shoved to the right: both bobs once again move over simultaneously with the pivot. Even when the bobs are not vertically aligned, as shown in Part G, the bobs move in conjunction with the pivot and without changing their angular orientation.Now you may be thinking that this is exactly what is needed to maintain a stable platform, but upon careful analysis we see that in fact we do not want a platform that have an infinite period of oscillation since then a horizontal displacement caused by a horizontal acceleration will mean that the platform is no longer locally horizontal once it is displaced from its initial location (remember the Earth's surface is curved). The infinite period pendulum was horizontal in its initial location, but once it has moved over it is no longer horizontal in the new location. Why is this a problem? Well, we have alluded to the fact that not having a locally horizontal platform means that the errors stemming from the need to integrate the accelerometer's output leads to an unbounded error in the position coordinate. On the other hand, when the platform maintains a locally horizontal orientation, then the positional error arising from integration of the accelerometer's output is bounded. We will examine just how this happens in greater detail in a moment. But for now, let's summarize: a 2 seconds oscillation period is too short, it does not keep the platform horizontal, while an infinite period is too long but for a different reason, the platform after the displacement is no longer locally horizontal. But perhaps this implies that for some value between 2 seconds and infinity there might be a period T_oscillation that is capable of keeping the platform locally horizontal? Let's see if this is indeed the case and if we can find this period.Puzzler 1: Consider the following experiment: Accelerate a Horizontal Platform, measure the acceleration, and use this information to drive the platform back to horizontal. Write the necessary equations for the motion, and compute the resulting period.
Do you recognize the period you computed in the first Puzzler query? Can you give a physical argument for why your calculation must have found this period?
Hint: Consider how long a simple pendulum would have to be in order to oscillate with this period.Puzzler 2: Why does a stable horizontal platform bound the drift errors of a navigational system while a platform allowed to move in 3 dimensions leads to an unbounded drift error?
Puzzler 2 is a much harder question to answer, but it is also a much more important result. It means that an inertial guidance system can be manufactured so that its navigational error is finite and bounded for all time. This means that the navigational system can be used for long periods of time without recalibration to a known location, for instance, out on the open sea where there are no known locations (there are no street signposts in the middle of the ocean, in other words). On the other hand, a 3 dimensional inertial system has an unbounded error and thus must be either a much more accurate and stable mechanism or it must be recalibrated to a known position more often.
Hint: Propagate the errors for both the locally horizontal stable platform and for the platform allowed to move in 3 dimensions.
Here is the solution PDF:This PDF file gives alternative physical explanations for the reason a pendulum having the proper period will provide a stable, locally horizontal, platform. It also derives the error analysis that shows that a locally horizontal platform leads to bounded drift errors while a platform in 3 dimensions allows for the drift errors to be unbounded.
(I will probably use this Puzzler as one of the FPF demonstrations.)
This problem, that of finding the curve along which a frictionless body will traverse in the same amount of time no matter how high is its starting point, is known as the tautochrone problem. The etymology of the word is that `chrone' stems from Kronos. In Greek mythology Kronos (Cronus) was one of the twelve Titans and father of Zeus who ate all of his children because it was prophesied that one of his sons would overthrow him. His wife Rhea fooled him, however, saving the baby Zeus to later become the supreme god. Kronos was confused with Chronos, the personification of time, and thus source of the root `chrone'. The prefix `tauto' is a contraction of the Greek to auto meaning `the same', as in tautology. Thus the word tautochrone means ``the same time''.
Tautochrone Curve - same time for all heights...
This problem has a long history, as it was initially realized in the 1600s that Galileo's pendulum is not truly isochronous, that the pendulum's period indeed depends slightly upon the amplitude of the swing. [Recall that the linearized pendulum, where the force is assumed linear instead of sinusoidal, is isochronous and thus for small amplitudes of the physical pendulum the period is nearly independent of the amplitude. But in the ``The Mathematics of Pendulums'' Section beginning on page 4203 of the Classical Mechanics Chapter we derived that the actual periods of the pendulum are given by Elliptic Integrals of the First Kind and their Jacobi Elliptic Functions.] Christiaan Huygens (1629-1695) set out to solve this nonisochronism problem and found a remarkable solution, that a pendulum whose bob traces out a cycloid instead of a circular arc would be isochronous. Thus Huygens built the first cycloidal pendulum in order to correct the slight nonisochronism of the physical pendulum, and cycloidal pendula were built and used throughout the 19th Century. Today, however, we do not find cycloidal pendula because it turns out that the friction produced by forcing the pendulum's bob to follow a cycloidal path generally leads to an error greater than the error stemming from the nonisochronism of the traditional pendulum whose bob follows a circular path, especially at small swing amplitudes.Now this problem, and the related brachistochrone (``shortest time'') problem, is often solved using the techniques of Calculus of Variations. Here we wish to put the Convolution Integral Theorem to good use, and thus we follow the strategy of Niels Henrik Abel's (1802-1829) solution.Since it is assumed that the body travels frictionlessly along the tautochrone curve, use conservation of energy to derive the equation controlling the motion of the body. Integrate this equation and set the time interval to a constant in order to determine the formula for the tautochrone curve. Then demonstrate that the tautochrone curve is a cycloid. How did Huygens construct an isochronous pendulum?<-- This PDF file solves the tautochrone problem using Laplace Transforms to simplify the integration. It also explains how Huygens solved the nonisochronism problem of a pendulum by requiring the bob to follow a cycloidal path.Museum Note: The Museo Galileo in Firenze, Italy has several nice exhibits demonstrating the tautochrone or brachistochrone curve.
This is a wooden troff, built in the 1800s, in the shape of a brachistochrone that illustrates that a ball will roll faster to a given vertical depth on a brachistochrone curve than on a straight incline.
This is an exhibit by the Panerai Italian watch company that illustrates that a pendulum bob following a tautochrone curve will outpace a pendulum bob following a circular curve. Photographs taken on July 20, 2012, Museo Galileo, Firenze, Italy.
This is a close-up of the cycloidal barriers restricting the pendulum bob's path to the tautochrone curve.
Or maybe you already think they should sound different? -- I mean, they are different lengths! But did you know that an 8' open organ pipe and a 4' closed organ pipe have the same pitch, that is, they sound the same fundamental frequency (roughly equivalent to the D2 note on a piano)? In which case, since they are sounding the same note, then why do they sound different?
These two organ pipes, while sounding the same note, do indeed sound different to the human ear. Why? [Thank you Justin Matters for demonstrating this on the Cathedral's organ to me and my progeny.]
The above diagram illustrates how a sound pulse propagates in a closed pipe.
Puzzler: By closely examining the difference between the above two diagrams, the one for the closed ended pipe and the open ended pipe, generate your own argument as to why a 4' closed pipe sounds the same fundamental note as an 8' open pipe. Then determine why these two pipes sound different even though they are at the same pitch.While this diagram shows how a sound pulse propagates in an open ended pipe.
Hint: Figure out the nodal structure of the standing waves in each pipe.
Here is the solution PDF:
The PDF file also explains why a pipe should make any sound at all, and why the flute and clarinet, both roughly 2' long, have such different bottom notes. It also explains why a Pressure Zone Microphone (PZM) has a 6dB greater sensitivity over a standard microphone.
In an effort not to leave anyone out, here is a problem for you Financial Gurus, Bankers, and CPAs...
Business, or financial, calculations are not difficult in principle, but they do require careful attention paid to details. Unfortunately, many business textbooks simply provide lists of formulae to employ along with directions on their usages. But a few of these textbooks do not provide the list of assumptions that go into deriving these formulae, and thus at times these formulae might inadvertently be applied inappropriately. In this chapter we attempt to provide the derivations of the financial formulae just so that the underlying assumptions are known. This rigorous proof approach also clarifies the nature of financial calculations, makes them easier to understand, and illustrates the connections between apparently different formulae, such as those for annuities and mortgages.
Financial computations are really nothing more than the calculation of interest on a sum of cash, sometimes the cash value of a commodity, at predetermined time intervals when both partial payments of principal and the interest are performed. Sometimes the interest is compounded, meaning that it is added to the principal for the next time period. [When interest is compounded at the end of each period, then the underlying mathematics becomes that of a geometric series.] Sometimes a balloon payment occurs which wipes out the debt before its schedule of payments has come to its normal conclusion. At times the commodity is depreciated, that is, its inherent value is presumed to decline over an assumed lifetime of its usefulness. The schedule of depreciation and the lifetime of the commodity depends upon the commodity itself, whether it is an automobile or a computer, for instance. And at times the interest is assumed to be compounded continuously, not just at the end of each time period. [In this case, the principal is an exponential function determined by a first-order differential equation.] And sometimes the payments are made at the start of the time period to which they apply while at other times the payments are made at the end of the time period to which they apply. In addition, the time periods may be irregular, not occurring after the same intervals of time, but this would be unusual as the time periods are typically and almost always assumed to be uniform. Also, sometimes the payment of principal portion may be suspended for a time with only the interest payment being made at the end of each period during the principal suspension. All of these potentialities produce complications for the calculation of the principal and interest repayments comprising an amortization schedule. And it is expressly for these various complications that any particular textbook financial formula might not be exactly applicable to any specific financial calculation, and it is for this reason that we derive the following financial formulae.
A Personal Vignette: Did you ever wonder if the amounts your lender charges you on your mortgage are correct? Believe it or not, I have personally changed the way Sallie Mae does business, by writing a 12 page letter to the President of Sallie Mae containing a rigorous mathematical proof of a particular amortization calculation when Sallie Mae claimed we owed more money ($498 more, to be precise) than what I had calculated myself. I tell this story, and give the rigorous proof, in the following PDF solution file. Moral of the Story: It is always a good idea to understand the underpinnings of any calculation so that you know what assumptions are being made in the computation. This keeps one from making a mistake like Sallie Mae made only to have someone call you on it.Here is a selection of the formulae derived:
- To compute the last payment period:
(Yes, those are "funny-looking" brackets -- they have a special meaning...)
- To compute the n-th Principal, valid for any n:
- For continuously compounding interest: [Differential equations are required.]
And a few of the problems solved are entitled:
By solving these problems and understanding the assumptions that are made, you too can check the calculations of your mortgage company... and the so-called Business Calculators are notorious for performing financial calculations without providing the assumptions inherent in those computations. Thus many errors have stemmed from using the incorrect calculation for a particular situation.
- Fixed Interest Rate Loan: Payment Amount
- Fixed Interest Rate Savings Plan: Monthly Savings Amount
- Fixed Interest Rate Loan: Last Period
- Fixed Interest Rate Loan: Final Balloon Payment
- Fixed Interest Rate Savings: Continuous Compounding Interest
- Fixed Interest Rate Savings: Interest Rate [This problem cannot be solved in closed form, thus its solution must be computed iteratively -- I provide several methods and the computer (Linux, Windows, Mac OS) programs to do so.]
- Fixed Interest Rate Bank Account: Final Principal
- Fixed Interest Rate Bank Account: Doubling Time
- Fixed Interest Rate Loan: Payment Amount and Total Interest
- Fixed Interest Rate Loan: Retired Early
- Fixed Interest Rate Lease: Payment Amount
Here is the Business Calculations Chapter from Math for the Motivated that includes all of these derivations and the solutions to the financial problems:Note: Just a comment on the pronunciation of words deriving from Latin... Now I don't really know how to pronounce these words, and, in fact, I don't think anybody really "knows" how to pronounce these words since Latin has not been continuously spoken since the times of Caesar. In fact, Latin has not been an actively spoken language by a culture for centuries, and thus the pronunciation of a Latin word is always somewhat of a guess. Frederic M. Wheelock (1902-1987) was one of the world's experts on Latin, and thus I always defer to the pronunciations given in Wheelock's Latin, recognized as being the premier comprehensive beginning Latin textbook. In particular, the ae in formulae is pronounced like the ai in aisle. You will find, however, other pronunciations of the word 'formulae', including other pronunciations given in various dictionaries. For instance, the Random House Dictionary gives the pronunciation of 'formulae' as 'f\^{o}r\prime\ my\schwa\ l\={e}', in direct contradiction to Wheelock's Latin. Over the years I have heard numerous different pronunciations of words like 'alumnae', 'alumnus', 'alumna', and 'alumni', but I do recall that in a commencement speech at Harvard University by Derek Bok (1930- ), the president of Harvard University from 1971 through 1991, that Derek Bok used Wheelock's pronunciations for these words and not the Random House Dictionary's pronunciations. Derek Bok's speech was given entirely in Latin --- how do you like that for a commencement speech? University Presidents are probably no longer required to speak Latin. (Having a little Latin under my belt, I did understand a modicum of Bok's talk, but at the time I wondered how many of the graduates listening to his oration understood a word. Fortunately, an English translation of the speech was passed out afterwards.)
[This is an Advanced Puzzler requiring knowledge of Differential Equations, Fourier Transforms, and Complex Analysis.]
An Undamped Harmonic Oscillator is modeled by the second-order ordinary differential equation:
We know that an Undamped Harmonic Oscillator oscillates as illustrated in the following Graph:
If we apply a periodic forcing function to this Undamped Oscillator, the amplitude of the oscillator's response depends upon the match between the frequency of the forcing function and the natural frequency of oscillation of the harmonic oscillator. As the forcing function's frequency approaches the natural frequency of the oscillator, the oscillator's amplitude peaks. For a truly undamped oscillator, the amplitude indeed goes to infinity, but all physical oscillators have at least a small amount of damping which keeps the amplitude finite. The following Graph illustrates this behavior known as resonance:
When the harmonic oscillator is damped, then we must add a "damping" term to the second-order differential equation:
And thus the Damped Harmonic Oscillator behaves as shown in the following Graph. Notice that it oscillates but the amplitude of the oscillation decreases exponentially over time:
When the Damped Harmonic Oscillator is also Forced, then it responds to the forcing function similar to the way the undamped oscillator responds, excepting now the resonance peak does not go to infinity but rather remains finite in amplitude:
The Forced Damped Harmonic Oscillator thus behaves as illustrated in the following Graph:
The Differential Equation modeling the behavior of a Forced Damped Harmonic Oscillator, when the forcing function is a cosine, is:
Click for full-size image
On the other hand, the forcing function might be any periodic function, g(t). If we then denote the partial derivatives using subscripts (u_{tt}=d^2u/dt^2, u_{t}=du/dt), we have for the arbitrarily forced, damped, harmonic oscillator:
When the forcing function g(t) is a sine or cosine wave, then this differential equation is easily solved. But what if g(t) is any arbitrary forcing function? What does the solution look like then?Puzzler: Solve the Forced, Damped Harmonic Oscillator for any Arbitrary Forcing Function.
Hint 1: Use the Fourier Transform and the Convolution Theorem to solve the Forced Damped Harmonic Oscillator differential equation. Use contour integration and the Residue Theorem of Complex Analysis to integrate the resulting integral for the Inverse Fourier Transform.
Hint 2: The following diagram illustrates the contour integral to use in order to compute the Inverse Fourier Transform integral along the real axis from -infinity to +infinity:
Here is the solution PDF file:
This PDF file also discusses how a data collection system composed of a physical transducer, a sample and hold circuit, an analog to digital converter, and a digital storage system may be analyzed using Fourier Transforms and the Convolution Theorem.
... and this PDF file,
derives the resonance behaviors for the various harmonic oscillators using the standard techniques, and it explains Nyquist Antialiasing and the Shannon Sampling Theorem, and what occurs when nonlinear perturbations are introduced...
What is the "crack" of a whip? Can you explain, from fundamental physical principles, why the bull whip cracks?
Hint: It has something to do with the way a wave is reflected off the end of the whip.
This PDF explains, in detail, the mechanism for the "crack":
What is the difference between percussion and music?
This is perhaps more of an opinion piece, in that I would call a musical instrument one that "sings" like the human voice and a percussion instrument one that does not "sing" in this fashion. With this personal "definition" of musical versus percussion, I would reclassify a piano as a musical instrument instead of as a percussion instrument, even though the piano strings are stimulated by a hammer in a percussive fashion.Puzzler: So, with this concept distinguishing percussion from musical instruments, how do they differ? (What is the difference between "singing" and not "singing"?)
Hint 1: The difference is related to the vibrational modes allowed in 1-dimension versus those allowed for a 2-dimensional circular membrane.
Hint 2: Consider Bessel Functions.
The distinction that, in my opinion, actually distinguishes percussion from musical instruments is described in the following PDF file:
Here is a photograph of one of the nodal patterns for a circular drumhead...
What is the physics behind the fact that in order to achieve your highest swinging motion you have to pump out-of-phase with the oscillation?
Puzzler: Give a Qualitative Explanation for Why One Must Pump 90 Degrees Out-of-phase in order to Achieve Maximum Swing Amplitude.
Hint: Consider what happens to a forced damped harmonic oscillator.Here is the solution:
Did you know that an insect smaller than the size of your pinky finger has caused five crashes of commercial jets?
Puzzler: What insect was it? And how did the insect cause a jetliner to crash? Explain the physics.
Hint: The answer has something to do with how Henri Pitot (1695-1771) measured the flow of the Seine River in Paris.
This PDF derives the solution and explains:
Very sad story: The Berginair Flight 301 (Boeing 757-225) crashed shortly after takeoff killing 186 people on February 6, 1996. The accident has been attributed to Sceliphron caementarium.
Say you have a glass tube whose diameters on its two ends differ. Blow two soap bubbles of the same size, R_1 = R_2, onto the ends of the tube. Are the bubbles stable?
In other words, what happens to the two soap bubbles in the above diagram? Some would say that since the hole is smaller at the right-hand end of the glass pipe, then the bubble on the right-hand end will grow smaller with the left-hand bubble getting bigger. Others would say that since the hole is larger at the left-hand end of the glass pipe, then the left-hand bubble will get smaller with the right-hand bubble growing in size. And still more would say that since the two bubbles are the same size then nothing will happen.Puzzler: What do you think?
And the answer is:
Even though the Coriolis Force appears to be a very small force that only manifests itself on the scale of hurricanes --- it is a very important aspect of our modern lives...Puzzler: Explain why the Coriolis Force is important to your ability to drive a car.
Hint: Recall how the torque converter works.
The following PDF file give what I call the grapefruit explanation of the fluid drive of a car. It also illustrates how the Coriolis force enters into flying an airplane, directing a spacecraft to Mars, taking a digital photograph, detecting molecules in interstellar space, plus more:
We drop a steel ball from a tall building, say from the 102-nd floor of the Empire State Building in New York City. We assume that there is no air resistance. Will it land "directly below" where it is hanging? And just what do we mean by "directly below"? (Neglect any effects from wind or air resistance.)
[The latitude of the Empire State Building is 40\degree 44' 54.36" N and the height of the 102-nd floor is 381 m. Since we don't know the actual gravity force at the Empire State Building, we will use g_0 = 9.80665 m/s^2. Furthermore, we have for the angular velocity of the Earth \omega = 7.292116x10^{-5} s^{-1} and for the Earth's mean radius R_{Earth} = 6.37103x10^{6} m. Calculate where the ball will fall.]
If the origin, O, of a coordinate system is directly below a plumb line hanging from the top of the Empire State Building, and the positive x-axis points directly East while the positive y-axis points directly North, then in which quadrant, I through IV, will the steel ball land?
Some people would say that the ball lands precisely where the plumb line points since the plumb line obviously hangs vertically and gives the direction of the force of gravity. Others would say that the ball lands somewhere along the positive x-axis since the 102-nd floor of the Empire State Building is actually heading to the East (since the Earth is spinning around its polar axis) faster than the building's first floor. [The top of the Empire State Building is farther away from the axis of rotation than the first floor, thus it's tangential velocity is faster.] When dropped, the ball moves East at a constant speed thus outrunning the easterly velocity of the ground causing the ball to land east of the plumb line. So, which is it?Puzzler: Does the dropped ball land at the origin O, along the +x-axis, or in one of the quadrants?
Hint: The ball neither lands at the origin nor along the +x-axis. See if you can calculate precisely where it lands. Remember the Coriolis Force.
And the answer is: . This PDF file includes an illustration of performing a perturbation calculation, that is, it goes beyond the typical Freshman Physics answer.
The recent Cycloramic app for the iPhone 5 was added to the iPhone App Store on December 21, 2012. It will spin the iPhone standing on its bottom edge while taking a panoramic video. Explanations of the mechanism, the actual physics, behind this spinning motion so far have been limited to just stating that the spinning is due to the vibration without any explanation as to how that vibration produces rotary motion. We collected a number of observations from various videos posted on the Internet, and we propose a physical mechanism that explains all of these observations (as of January 14, 2013). Below are the observations, what mechanism do you propose that satisfies these observations?
Citations for the following observations are included in the CM-Gyroscope_CycloramicApp_pw112.pdf PDF file.
Looking at a number of YouTube videos as well as the videos on the cycloramic.com website, we notice a few things. First of all, nobody, including the app's designers, have a physical explanation of how this app causes the iPhone to rotate. Everything that we could find only states that the app uses the vibration of the iPhone in order to rotate, but nobody says anything beyond this. Of course, simply stating that it is the vibration of the iPhone that causes the phone to rotate is not a proper physics mechanism explaining how and why it does so. Now there are a few principle observations that we found in various videos (see the CM-Gyroscope_CycloramicApp_pw112.pdf for the references) and on various reviews of the Cycloramic.app, listed below:
- Most obviously, the iPhone 5 rotates in a single direction when standing on its bottom edge with its vibration ringer operating.
- The Cycloramic.app operates on the iPhone 5 but the developer states that it does not work on the iPhone 4/4S. Several individuals have left comments on YouTube demonstration videos stating that the Cycloramic.app does not work on the iPhone 4/4S.
- The iPhone 5 uses a vibration rotary motor spinning an eccentric cam as its generator of vibration.
- The iPhone 4/4S uses a linear oscillating vibrator, similar to a speaker cone and coil, as its generator of vibration.
- One individual in a YouTube video states, but does not show in the video, that the iPhone 4S rotates only a partial way around a circle but not a full circle. Another app, the iTelekinect.app, will move but not rotate an iPhone 4S.
- The Cycloramic.app will rotate an iPhone 5 even with the olloclip fisheye adapter attached, thus increasing the weight of the iPhone.
- In another demonstration video, the iPhone is resting on a small and thin piece of glass sitting on top of a smooth wooden board. When the Cycloramic.app is started, the iPhone barely moves, but when the individual presses down on the glass so that it is held securely to the wooden board, the iPhone does rotate.
- The ``Cycloramic App Hack Improves iPhone 5 Performance'' video demonstrates how a small, four-thicknesses, piece of Scotch transparent tape applied to the bottom of the iPhone 5 improves its spinning performance.
Puzzler: Given these observations, see if you can propose a physical mechanism that explains them all.
And the answer is provided on the wiki page: Gyroscope Precession: Cycloramic.app
There are very few 0s, 100%s, and infinities in Physics. There is very little perfection in Physics. Most everything has some friction, some leakage, some entropy, some heat, some temperature, some randomness that destroys the ideal behavior that one might wish to occur. Now there is the zero resistance of superconductors, the zero viscosity (no friction) of superfluid helium, the perfect diamagnetism of superconductors, the 100% reflectance of total internal reflection, the infinite density of a black hole (something you probably don't want to experience), and here is a mechanical device whose vibration frequency goes to infinity! Yup, you heard correctly, a mechanical device that vibrates at an infinite frequency! And you get to hear it in operation (briefly)...
(Work still in Progress...)
Foucault not only invented the gyroscope, measured the speed of light, and tinkered with early photography, but he was also the first person to experimentally demonstrate that the Earth turns?
(Work still in Progress...)
Copyright (c) 2010-2016 Craig G. Shaefer, all rights reserved.
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