1 USEFUL HINTS FOR SOLVING PHYSICS OLYMPIAD PROBLEMS By: Ian Bloland, Augustana Capus, University of Alberta For: Physics Olypiad Weeend, April 6, 008, UofA Introduction: Physicists often attept to solve difficult probles with siple ethods. There are several reasons for doing this: It s faster: the siple ethod provides a quic approxiation to a ore coplete solution that could be obtained with a ore coplicated ethod. You have no choice: the proble ight be too difficult to solve without a siplification. Another perspective: the proble has already been solved and we want to solve it again with a siple ethod, both to chec our answer and to see if we can gain soe additional insight into the solution. What I d lie to tal about today are soe of the siple ethods that you ight use to solve physics probles. In particular, there are three ain ideas that I would lie to discuss: (1) Diensional analysis () Oscillations (3) Approxiations Soe of this aterial was also given in a siilar tal at the Physics Olypiad Weeend at UBC in 006 and the transcript of that tal is also available on the web. First Exaple: Consider the vertical oscillations of a ass hanging fro the end of a spring. Without a direct easureent of the oscillation period, is there a way that we could predict this? Sure, there ust be soe sort of forula: T =? Even if we don t now what this forula is, it turns out that we can guess it. The first step is to deterine the variables that T ight depend on. It sees reasonable to expect that T ight depend on as any as four variables:, the ass which hangs fro the spring;, the spring stiffness constant; A, the aplitude of the oscillations; and g, the acceleration due to gravity.
In the second step, we guess that the forula involves only ultiplication and division of various powers of our variables, α β γ δ T = C A g where C is a diensionless constant and α, β, γ, and δ are unnown exponents. The third step involves exaining the units which appear in our forula. All correct equations in physics have consistent units, therefore we require We can replace the N with α [ s] = [1][g] N β [] g /s (thin of F = a [s] = [g] s s This leads to three equations, one for each unit: s 1 = β δ γ δ s ) so that we have β δ α g γ α + β γ + δ β δ [] = [g] [] [s] g 0 = α + β 0 = γ + δ At this stage, we have a proble: 3 equations involving 4 unnowns do not lead to a unique solution, so at this stage we would conclude that there are any possible ways to write a forula for the oscillation period that has consistent units. Suppose we do a quic bit of experientation to see how the oscillations change if the spring is not vertical. It can quicly be seen that the oscillations have the sae period if the spring is vertical, horizontal, or at any angle in between. This eans that the oscillation period does not depend on g and therefore that δ = 0 in the equations above. This leaves us with 3 equations for 3 reaining unnowns, s 1 = β g 0 = α + β 0 = γ These are easily solved: β = 1, α = β = + 1, and γ = 0. Substituting these exponents into the original forula for T, we have 1/ 1/ 0 0 T = C A g = C This technique of guessing a forula by seeing if there is a unique way for the units to wor out is nown as diensional analysis. In the exaple considered here, we have shown that the oscillation period cannot depend on the aplitude of the oscillations. This technique cannot tell us about the diensionless constant C, however. Only with an explicit derivation using the laws of physics can we deterine that C = π so that T = π
3 Explicit Derivation: Where does the π coe fro in the forula for the oscillation period? We ll loo at two different ways of showing this. The first ethod involves calculus, and while calculus is not required to solve Physics Olypiad probles, it is allowed. The starting point is Newton s second law, Σ F = a. According to Hooe s Law, the spring exerts a force F = x on an attached ass, where is the spring stiffness constant, x is the displaceent of the ass fro the equilibriu point, and the inus sign indicates that the force will always be directed bac towards the equilibriu point. We will ignore gravity here, since it erely changes the exact location of the equilibriu point without affecting the oscillation period. This leads to x = a and writing acceleration as the second derivative of displaceent, d x + x = 0 dt The otion of the ass is given by the solution of this differential equation: x ( t) = Asin t (This can be checed by differentiation. Note that cosine could also be used.) Woring in radians, the sine function repeats itself with a period of π, therefore the otion of the ass will repeat itself after a tie T given by the condition T = π T = π Without calculus, we need to now that the Hooe s Law force F = x is associated 1 with a potential energy U ( x) = x. (Technically this result is also obtained via calculus, as U ( x) = F( x) dx, but this particular forula for U (x) can be found as the area of a triangle under the graph of F(x) vs. x.) The total energy of the oscillating spring-ass syste inetic and potential is conserved, therefore 1 1 v MAX = A since on the LHS, all of the syste s energy is inetic at the equilibriu point while on the RHS, all of the syste s energy is potential at the turnaround points. Next, we recognize the one-diensional oscillations of the ass on a spring as the shadow of two-diensional unifor circular otion whereby the consideration of one coplete circle leads to πa A vmax = T = π T v The energy conservation equation allows us to replace the ratio A / vmax with / so that we once again obtain the correct forula for the oscillation period: T = π MAX
4 A Useful Approxiation: In anticipation of the next exaple, let s step away fro physics in order to construct a very iportant atheatical approxiation. Consider the following siple algebraic identities: 1 + x = 1 + x + x ( ) 3 3 ( 1 + x) = 1 + 3x + 3x + x 4 3 4 ( 1 + x) = 1 + 4x + 6x + 4x + x If x is uch less than one (or, if x is negative, x << 1) then x is saller still and it ight not be a bad approxiation to ignore it (and even higher powers of x) altogether. This leads to 1 + x 1 + x The pattern is obvious: ( ) 3 ( 1 + x) 1 + 3x 4 ( 1 + x) 1 + 4x ( 1 + x) n 1 + nx for x << 1 It turns out that this approxiation is valid for any value of n, not just whole nubers. This is nown as the binoial approxiation. Second Exaple: Perhaps the ost failiar oscillating syste is the siple pendulu, where a sall bob of ass swings at the end of a light string of length L. As an exercise, you can confir that diensional analysis on the variables L, g, and leads to the forula T = C L / g. We will now obtain this result in a different way. Since the pendulu bob oves bac and forth in arcs of a circular path, we can easily setch the potential energy function U ( x) = gy( x) by taing the botto half of the circle x + ( y L) = L so that U ( x) = g L L x
5 Near the botto, the circle can be approxiated by a parabola: 1 This eans that the potential energy function is approxiately U ( x) = x when x is near zero. If we can deterine the value of which provides the atch, we will be able to deterine the period of the pendulu fro T = π /. We do this by algebraically approxiating the exact expression for the potential energy: U ( x) = g L L x = gl 1 gl 1 ( x 1 ( 1 { 1 ( x / L )}) x gl L 1 g x L 1 1 Note that the binoial approxiation (with n = and " x" = ( x / L ) ) was used in the third line under the assuption that ( x / L ) << 1. This allows us to identify the value needed to find the period of the pendulu: g L = T L = π π ( g / L) = g / L ) General Oscillations: The ethod used in the previous exaple is applicable to any ind of oscillating syste, particularly when the oscillations have a sall aplitude. To reiterate, we are looing to find a value for the oscillating syste, using either (1) forces, F = x, or () potential energy, U = 1 x, where x is the displaceent of the syste fro a point of stable equilibriu. Once we have deterined the value, the oscillation period is siply T = π /.
6 Third Exaple: Consider a large glass jar with a tall, thin tube projecting upwards fro the top (see the setch below). If a ball bearing, whose diaeter is such that it just barely fits into the tube, is dropped into the tube, gravity will pull the ball down the tube. Then a strange thing happens: the ball slows down, stops, and bounces bac up towards the top of the tube whereby it continues to oscillate up and down. y This occurs because air cannot go around the ball fro one side of the tube to the other, therefore the air in the jar is copressed as the ball falls. Air pressure inside the jar is increased as a result and this leads to an upward force on the ball. If we denote the ball s position in the tube by a coordinate y which is zero at the top and points upward, it can be shown that the net force acting on the ball is γpat A F( y) = y g V where: γ 1. 4 is a constant which relates to the details of how easily air can be copressed; P at is atospheric pressure; A is the cross-sectional area of the tube; V is the volue of air in the jar; and is the ass of the ball. This force equation is of the for F = y y ), therefore we can deterine the period of the oscillations: ( 0 γpat A = V T = π = π γp V at A Fourth Exaple: In the fourth question of the 008 selection exa, an electrically charged dust particle was suspended above an electrically charged ring:
7 F e g h Based on the net force Qqh F( h) = g 3 / ( r + h ) it was deterined that there are two equilibriu points where F = 0 : the first is close to the ring and is unstable while the second is further fro the ring and is stable (in the vertical direction). This is shown in the graph below (where x = h / r ). If the particle is at the stable equilibriu point and is gently nudged, it will oscillate about the equilibriu point. Writing h = ( h h + h = y + h EQM ) it is left as a (challenging) exercise to show that the net force can be written as Qq(hEQM r ) F( y) y 5 / ( r + h ) For the stable equilibriu, where (h r ) 0, this leads to an expression of the for EQM > F = y (where this latter is a spring constant, not to be confused with Coulob s constant in the earlier force expressions in this exaple) that can be used to deterine the period of oscillations of the dust particle. For the unstable equilibriu, (h r ) 0 EQM EQM EQM EQM < and the force law becoes F = + x and no oscillations occur; the dust particle is pushed farther and farther fro equilibriu instead.
8 Suary: I have outlined three general techniques here that can be extreely useful for solving physics probles: (1) Diensional analysis Every equation in physics ust have consistent units. This obvious but powerful stateent allows us to spot istaes and, in soe cases, to guess the correct for of an equation based on the units of the variables it will depend on. () Oscillations The period of alost any oscillating syste can be expressed as T = π / if we can deterine a value either fro a force law lie F = x or a potential energy of the for 1 U = x. (3) Approxiations Often we are willing to trade an exact solution for an approxiate one if it leads to a siplification in an equation. In particular, the binoial approxiation ( 1 + x) n 1 + nx, valid when x << 1, is a very useful thing to now. Bonus: The following Fox Trot cartoon contains an error in one of the equations. Can you spot it?