A Model for Collisions Between Two Solid Objects One way mechanical energy is lost during collision between two solid objects is by way of acoustic waves propagating inside both objects. Even though the real situation is quite complicated, in this problem it is modeled in a simple way. First treat each solid rod as a spring with unstretched length L l and and L L r . Spring constant times the length of the spring are then K then K l and and K K r , respectively. Mass density (mass per unit length) of the spring are given by ρ by ρ l and and ρ Indexes l and r represent the left and right springs. ρ r . Indexes l The left spring is moving with velocity +v + v0 /2, while the right spring is moving in the opposite direction with velocity −v0 /2. Each spring is initially relaxed. At t At t = 0, the springs collide at x at x = 0. The displacement of each point on the springs are described by a function y function y((x, t) so that x + y + y((x, t) represents the position of said point at time t; t ; this point was at x at x when t when t = 0. 1. Derive Derive the wa wave ve equation equation on the springs and write down the speed of the wave. wave. The general solution to the wave equation is given by y( + φ((ct + ), where y (x, t) = ψ( ψ (ct − x) + φ ct + x x), of ψ and and φ c is the speed of the wave propagation. The form of ψ φ are determined from the boundary conditions. 2. Write down down the boundary conditions conditions at x = 0, x 0, x = =
−Ll
and x and = L x = L r .
3. Write down down the function of y ( t ≤ 0), i.e. y0,l (x, t) and y (x, t) before the collision (t y0,r (x, t). At t At t = = 0, an acoustic wave starts to propagate in both springs away from the collision point x = 0. The dynamics of the system is analyzed using the space-time diagram as shown in Fig. 1. The horizontal axis represents time, and the vertical axis represents the position of points on the spring. Each line in the diagram represents an acoustic wave front that emerges each time a wave front arrives at the border. For example, the line AB represents a wave front emerging from the collision at point A (x = 0). Waves propagating in the left and right springs are given by functions f l (cl t + x + x)) and f and where c l and and c f r (cr t − x), where c c r are the speed of the wave propagation in left and right spring, respectively. The space-time diagram indicates that L l /cl > Lr /cr in this problem. As the wave front of f of f r (cr t − x) arrives at point B, a new reflected wave, gr (cr t + x + x), ), emerges. The same event also occurs in the left spring at point C. Now back in the right spring, when the wave front of gr (cr t + x + x)) arrives at the end of the spring (x (x = 0, at point D in the diagram), a new reflected wave h r (cr t − x) and a new transmitted wave wave h l (cl t + x + x)) are generated. These phenomena always occur when a wave front arrives at the border; a new reflected wave or new reflected and transmitted waves are generated. 4. Write down the wave wave function y function y((x, t) in the region I, II, III, IV, V, VI and VII in terms of y0 , f r , f l , g r , h r and hl .
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Figure 1: Space-time diagram 5. Using the boundary condition(s), determine the form of f r (cr t − x) and f l (cl t + x) in terms of the springs’ properties and initial velocity. 6. Determine the velocity of the contact point ( x = 0) immediately after the collision. 7. Using the boundary condition(s), determine the form of g r (cr t + x) in terms of the springs’ properties and initial velocities. Now consider a case where both springs are identical except in its length. In this case, ρl = ρr = ρ, K l = K r = K . Take L r < L l . 8. Determine y(x, t) in region III and IV. Draw a graph for y(x) at t = 0.4 Lc . For drawing the graph, you may use L r = 0.6L, L l = L and v 0 = 0.5c. 9. Determine y(x, t) in region V. Draw a graph for y(x) at t = 0.8 Lc , use the same L r , L l and v 0 as in the previous question. 10. When will the two springs separate? Draw a graph for y(x), use the same L r , L l and v 0 as in the previous question. 11. Calculate the coefficient of restitution e between the springs. 12. Calculate the ratio of the translational kinetic energy of the springs after the collision to the kinetic energy before the collision. Page 2 of 2
Spring and Mass Problem 1. A mass M moves toward a semi infinite spring with initial velocity v0 , as shown in Fig. 1. The spring has mass per unit length µ and spring constant times the spring length K ≡ kL. The mass and the spring collide at x = 0 and t = 0. Write down the velocity of the mass M after the collision as a function of time, and also write down the velocity of the mass M as a function of position.
Figure 1: 2. For this part another mass m is placed at the other end of the spring. After the initial wave front from the collision of mass M with the spring reaches this mass, how long would it take for this mass to leave the spring? Also calculate the velocity of mass m when it leaves the spring. Assume the waves in the spring travel faster than the initial velocity of mass M but the spring is long enough so that when the mass m leaves the spring, the reflected waves from m have not yet returned to M .
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Sliding Block A rectangular block of width 2b, length 2a, and mass M rests on a rough surface which has a coefficient of kinetic friction µ. At some time, the block is given a sharp kick, such that it suddenly attains a horizontal velocity v0 . Under certain circumstances the rear end of the
Figure 1: The block, after given its initial velocity. block will begin to lift and the block will subsequently rotate about its front lower edge, which will remain in contact with the surface. 1. Derive the equation of rotational motion of the block in terms of θ, a, b, µ, and g. 2. Find the physical condition, namely the range of µ, that allows this to happen. The next question assume this condition is fulfilled, and concerns the subsequent motion of the block. 3. Consider a final state in which the block is at rest in the position shown in Fig. 2, where its center of mass has undergone a total horizontal displacement x. Is such a
Figure 2: A presumed final position of the sliding block. position possible? If yes, calculate the initial velocity required to achieve it for the following values: a = 0.8 m, b = 1.0 m, µ = 0.9, x = 1.65 m, θ˙max = 1.27 s−1 . Note: Knowing a, b and µ the initial velocity can be solved numerically.
Dielectric Slab Waveguide 1
Total Internal Reflection
The electric field of a polarized monochromatic plane wave can be generally represented as E(r, t) = E exp i(k.r ωt), where E is the amplitude of the wave, k the wavenumber, and ω the frequency. Suppose that a monochromatic plane wave with frequency ω travels in the medium of refractive index n1 , and is incident on the boundary of another medium of refractive index n2 . The incoming wave forms an angle θi with respect to the normal of the boundary. Throughout this problem, we only consider transverse electric (TE) polarized wave where the electric field is perpendicular to the plane of incidence and all media are non-magnetic.
−
1. In the case of n 1 > n2 , there exists a critical angle θ c where the incoming wave will be totally reflected for θ i > θc (total internal reflection or TIR). The phase of the reflected wave lags by δ with respect to the incident wave. Derive δ and state it in terms of n 1 , n2 , and θi . 2. Using the necessary boundary conditions, derive the reflectance R for the case of TIR. Show that the wave is perfectly reflected for all θ i > θc.
2
Constructive Phase Matching
The most simple dielectric waveguide is a planar slab with thickness d and refractive index n1 located in a homogeneous background medium with refractive index n 2 (n2 < n1 ). In the case of TIR, the slab can be used to guide waves without loss, with the additional condition that the waves interfere constructively. In other words, the wavefronts should be preserved as the waves travel inside the waveguide. The wavenumbers for the vacuum, medium n1 , and medium n2 are taken to be k 0 , k1 , and k2, respectively.
1. Find the necessary condition for the constructive phase matching. 2. The wave can only be guided without loss for certain values of θ. Show that in these cases, θ must satisfy the equations: k1 d cos θ
− δ = mπ;
m = 0, 1, 2, 3,....
(1)
Verify that the equations above can also be written as:
√
k0 d u + v = 2 2
2
u tan u = v with u =
3
k1 d 2
cos θ and v =
d 2
k
2 1
or
sin2 θ
2 2
n21
2 2
− n ,
− u cot u = v,
(2) (3)
− k .
Maxwell’s Equations
The Maxwell wave equation for the electric field in a dielectric medium of relative permittivity ε is ∂ 2 ∂ 2 ∂ 2 ∂ 2 E(r, t) E(r, t) = µ 0 εε0 + + (4) . ∂x 2 ∂y 2 ∂z 2 ∂t 2
In the case of the slab waveguide shown in the figure above, ε = n 21 for 0 < z < d, and ε = n 22 for z < 0 or z > d. Taking the system coordinates such that the wave travels in the xz -plane, the electric field can be generally written as E(r, t) = E (x , z , t) = E (z ) exp i(βx
− ωt),
(5)
where β is the effective propagation constant along the waveguide due to the translational symmetry of the structure in the x-direction. In the case of waveguiding the TE polarized ˆ), E(r, t) should be simple harmonic inside the slab and decay wave (E(z ) = E (z )y exponentially outside. 1. What is the relation of β to k 1 and θ? 2. From the boundary conditions at z = 0 and z = d, derive from the Maxwell equations the condition for waveguiding as found in Part 2. 3. Sketch the functions in eqs. (1)-(2) in ( u, v) coordinates. Determine the necessary condition for only one solution of θ to exist.
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4
Mode Solutions
The waveguide mode solutions are solutions of θ where waveguiding occurs inside the slab. The solution for m = 0 (see Part 2) is commonly called the fundamental mode (the lowest mode or the first mode), the m = 1 mode is called as the second mode, and so on. 1. Sketch the functions in eqs. (2)-(3) in ( u, v) coordinates. Determine the necessary condition for only one mode solution to exist. 2. Show that the maximum number of modes supported by the dielectric slab is M =
k d 0
n21
π
− n
2 2
,
(6)
where the symbol denotes the ceiling function for which the expression inside is increased to the nearest integer.
3. Verify that the number of mode solutions is incremented by one for every increase of frequency: πc ∆ω = (7) . d n21 n22
−
4. From eq.1, show that the group velocity ( dω/dβ ) of each supported mode solution is vg =
∂δ ∂β ∂δ ∂ω
d tan θ + n1 d c cos θ
−
.
(8)
5. Show that the maximum time disparity for different modes in the dielectric slab waveguide to travel a distance L is L τ = (n1 c
− n ). 2
6. For n 1 = 1.7, n2 = 1.5, λ = 800 nm (in vacuum), and d = 1 µm, find all the mode solutions for θ (with θ > θc ). Plot the electric field E (z ) for these solutions.
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(9)
Liquid Air A mixture of oxygen and nitrogen gas is stored in a closed container equipped with a piston on one end at a temperature of T = 77.4 K. The total amount of the gas mixture is 1 .1 mole and its initial pressure is 0 .5 atm. With the help of the piston the gas mixture is slowly compressed at constant temperature. Using plausible assumptions, plot the pressure of the system as a function of its volume until one tenth of the initial volume, if the ratio of the number of moles of oxygen to the number of moles of nitrogen is a) b) c)
nO2 nN2 nO2 nN2 nO2 nN2
1 . 9 2 = . 9 1 = . 4 =
Find the pressure and volume at distinctive points of these isothermal curves. You can use the following data: •
Boiling point of liquid nitrogen at 1 atmosphere: 77 .4 K
•
Boiling point of liquid oxygen at 1 atmosphere: 90 .2 K
•
Heat of vaporization of oxygen: 213 J/g.
Lagrange Points Stability In a system that rotates with the Earth around the Sun, there are five equilibrium points (where the sum of the forces is zero). These 5 points are known as Lagrange Points (named after Joseph Lagrange, the first person to study this three-body system). Exact analysis of this system is very complicated and chaotic. In the following problem, the mass of the two bodies (M 1 and M 2 ) are taken to be much larger than that of the third body ( m). The distance between M 1 and M 2 is taken to be R.
1. Basic equations of the system (a) Write down the vector of the total gravitational forces Fg on m. (b) By assuming M 1 , M 2 >> m, determine the angular velocity of the M 1 and M 2 system (Ω). (c) In a frame that rotates with the system, there are fictitious forces on m. Write down the vector of the total forces on this mass ( FΩ ) in this frame. (d) Choose a coordinate system where the three masses are in the xy-plane and the angular velocity Ω is in the positive z -axis. The center of the coordinate is set at the center of mass of M 1 and M 2 on the x-axis. Write the position of m as ˆ In the rotating frame where m is at rest, write down the total r = x(t)ˆi + y(t) j. forces on m in the x- and y-axis using parameter α = M M and β = M M . +M +M 2
1
1
2
1
2
2. Identifying Lagrange Points There are 5 points with zero net forces in this rotating system. Three of them (call them L 1, L 2 and L 3 ) lie on the line connecting M 1 and M 2 (the x-axis) and the other two (call them L4 and L5 ) lie on the xy-plane on symmetric positions above and below the x-axis; that is, y 4 = −y5 .
(a) First consider the case of finding the position of L 1 , L 2 and L 3. Use x = (ν − α)R, with ν the distance of m from M 1 in units of R. Write down the equation of force that must be satisfied to identify these points. Express this equation in terms of ν and α. (b) The equation above gives rise to three cases (each for L 1 , L 2 and L 3) to consider, ν < a, a < ν < b and b < ν . Determine the values of a and b. From here on, we will also assume that α is small (in the Earth-Sun system, α is 3.0 × 10−6 ). Use only the lowest order non-zero term in α, ignore all higher order terms in α. The following three questions will help you determine the three Lagrange points on the x-axis. (c) For the first case, ν < a, write ν = −1 + δ 1 with δ 1 a small positive number that depends on α. This value of ν will determine the position of the first Lagrange point at x = −R(1 + ξ 1 ). Determine ξ 1 as a function of α. (d) For the second case, a < ν < b, write ν = 1 − δ 2 with δ 2 a small positive number that depends on α. This value of ν will determine the position of the second Lagrange point at x = R(1 − ξ 2 ). Determine ξ 2 as a function of α. (e) For the third case, b < ν , write ν = 1 + δ 3 with δ 3 a small positive number that depends on α. This value of ν will determine the position of the third Lagrange point at x = R(1 + ξ 3 ). Determine ξ 3 as a function of α. Determining the fourth and fifth Lagrange points requires a more complicated method. First decompose the gravitational force on m into components parallel and perpendicular to the vector r. (f) Find the unit vector parallel to the vector r, ˆe . Find also the unit vector perpendicular to the vector r on the xy-plane, ˆe⊥ . (g) Find the component of the force on m parallel to the vector r, F Ω , and find the component perpendicular to the vector r, F Ω⊥ . (h) Specify the condition that must be satisfied by the force component perpendicular to the vector r in order that mass m be in equilibrium. With this condition, determine the relation between r m1 and rm2. (i) Specify the condition that must be satisfied by the force component parallel to the vector r in order that mass m be in equilibrium. With this equation, determine the relation between rm1 and R. (j) Now determine the position of the fourth Lagrange point (x4 , y4 ) and the fifth Lagrange point (x5 , y5 ). 3. Lagrange Point Stability To test the stability of these Lagrange points, small perturbation are given to the mass m around its equilibrium points. Because the forces in this system depend on the Page 2
position (x, y) and the velocity (vx , vy ) of the mass m, the restoring forces must be calculated for variations in position and velocity. Expand the total force as follows: ∂F x ∂ F x ∂ F x ∂ F x δx + δy + δvx + δvy ∂x ∂y ∂v x ∂v y ∂F y ∂ F y ∂ F y ∂ F y F y (x0 + δx, y0 + δy, vx,0 + δv x , vy,0 + δv y ) = δx + δy + δvx + δv y . ∂x ∂y ∂v x ∂v y
F x (x0 + δx, y0 + δy, vx,0 + δv x , vy,0 + δv y ) =
This force has taken into account the contribution of the velocity of the mass m. All the partial derivatives are evaluated at the point ( x0 , y0 , vx,0 , vy, 0). (a) Write down the general form for (b) Calculate m1 ∂F , m1 ∂F , m1 ∂F , ∂v ∂v ∂v the five Lagrange points. x
x
y
x
y
x
1 ∂F x m ∂x
,
1 ∂F y m ∂v y
1 ∂F x m ∂y
,
1 ∂F y m ∂x
,
1 ∂F y m ∂y
. Show that
∂F y ∂x
=
∂F x ∂y
.
. These four coefficients should not change for
These eight coefficients should act as a restoring constant (analog to the spring constant). Now we are ready to check the stability of the five Lagrange points. Consider only the lowest order term in α, ignore all higher order terms. (c) The first Lagrange Point i. Show that m1 ∂F = c 1Ω2 . Determine c 1. ∂x x
ii. Show that iii. Show that
∂F y ∂x 1 ∂F y m ∂y
=
∂F x ∂y
= 0.
= c 2 αΩ2 . Determine c2 .
iv. By substituting δx = Ae λt and δy = Beλt , with A and B nonzero, determine λ as a function of α and Ω only. v. There are four solutions to λ. Write down the condition that these solutions must satisfy in order that the first Lagrange point is stable and then determine the stability of this point. vi. For the Earth-Sun system α is 3.0 × 10−6 and Ω is 2π/year. If this point is stable, determine its period of oscillation (in days), if not, determine its time constant 1/λ (in days also). (d) The second Lagrange Point i. Show that m1 ∂F = c 3Ω2 . Determine c 3. ∂x x
ii. Show that iii. Show that
∂F y ∂x 1 ∂F y m ∂y
=
∂F x ∂y
= 0.
= c 4 Ω2 . Determine c 4 .
iv. By substituting δx = Ae λt and δy = Beλt , with A and B nonzero, determine λ as a function of α and Ω only. v. There are four solutions to λ. Write down the condition that these solutions must satisfy in order that the second Lagrange point is stable and then determine the stability of this point. Page 3
vi. For the Earth-Sun system: if this point is stable, determine its period of oscillation (in days), if not, determine its time constant 1 /λ (in days also). The third Lagrange point is similar to the second Lagrange point hence it need not be considered. (e) The fourth Lagrange Point i. Show that m1 ∂F = c 5Ω2 . Determine c 5. ∂x x
ii. Show that iii. Show that
1 ∂F y m ∂x 1 ∂F y m ∂y
=
1 ∂F x m ∂y
= (c6 + c7 α)Ω2 . Determine c6 and c 7 .
= c 8 Ω2 . Determine c 8 .
iv. By substituting δx = Ae λt and δy = Beλt , with A and B nonzero, determine λ as a function of α and Ω only. v. Define M 1 /M 2 = ξ . Find the range of value of ξ for the fourth Lagrange point to be stable. The fifth Lagrange point has the same behavior as the fourth Lagrange point, hence it need not be considered.
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Helical Rope A device is built by placing two metal strips, each of mass m, on a large, frictionless cylinder. Two identical massless elastic ropes each with spring constant k and obeying Hooke’s Law are used to connect the metal strips such that the two ropes are initially at their natural length x0 and parallel to each other. The contact points of each rope on the same strip are diametrically opposite to each other, and the whole device can be seen on Figure 1. Strip A is bolted to the cylinder, while strip B is free to move along and rotate about the cylinder’s axis.
Figure 1: The device in its initial configuration. The screw can be used to prevent rotation of strip B. 1. The cylinder is now oriented such that its axis is vertical under a constant gravitational field g, and strip A is above strip B. Strip B is now rotated N times while its distance from strip A is maintained at x0 . After this, strip B is prevented from rotating by a screw, as shown in Figure 1. (a) Find an equation that, given numerical values for the initial conditions, would allow you to calculate x1 , the new equilibrium position. (b) Under certain conditions the metal strip will undergo simple harmonic oscillation. Calculate the frequency of oscillation for small ∆ x, in terms of k, r, N , x 0, and x1 . 2. The cylinder is now oriented horizontally, the ropes are returned to their initial position, and strip B is prevented from rotating by means of the screw. (a) A horizontal stretching force F is now applied to strip B. If the force is increased very gradually, the ropes break when the force reaches a value F 0 . What, then, is the minimum amount of constant force required to break the ropes? (b) If strip B is rotated N times before the screw is put in place keeping the total length of the ropes x0 , calculate the minimum horizontal force required to break the ropes if said force is: i. increased very gradually. ii. kept constant.
3. The system is returned to its initial horizontal configuration. All constraints on strip B are now removed, and the strip is rotated through an angle θ0 while keeping the distance between the two metal strip x0 , and then released (initially x˙ (0) = 0, and θ˙(0) = 0). (a) Find the equation of motion of strip B! (b) Solve the equation of motion for x (t) and θ (t)! (c) Find the maximum velocity and maximum angular velocity, and also the time T required for strip B to reach strip A!
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Magnetic Dipole Oscillation A magnetic dipole with magnetic moment m 1 is placed at the coordinate origin parallel to the x-axis. 1. Determine the resulting magnetic field in all space. 2. Another dipole is placed at a distance r from the origin at an angle θ to the x-axis. The magnetic moment of the second dipole, m 2 , forms an angle α to the x-axis. The whole set-up can be seen in Fig. 1. Determine the torque on the second dipole. 3. Determine the interaction energy between the two dipoles. 4. Determine the force on the second dipole. 5. The second dipole is tied to the first dipole via a massless string such that the distance between the two is fixed at R. While the orientation of the first dipole at the coordinate origin is fixed, the orientation of the second dipole may change. It is also allowed to move freely in the xy -plane around the first dipole. Write down the equation of motion of the second dipole. The mass and moment of inertia of the second dipole are taken to be m and I respectively. 6. Initially the second dipole is at rest on the x-axis, with the magnetic moment forming an angle α0 to the x-axis (α0 << 1). At t = 0, the second dipole is released and allowed to move freely. Write down the equation of motion of the second dipole assuming θ and α are small. 7. The system undergoes simple harmonic oscillation. You are asked to determine the normal mode frequencies of oscillation. The system is in a normal mode when the oscillating variables are in phase and can be written as follows: θ = θ 0 cos (ωt + φ) and α = α 0 cos (ωt + φ). There are two possible values of ω (denoted by ω1 and ω2 ). Determine ω 1 and ω 2 . 8. For each normal mode, determine the ratio of the amplitude of α to θ, c1 = α1 /θ1 and c2 = α2 /θ2 . 9. The equation of motion of the system can be expressed as follows: θ = θ1 cos (ω1 t + φ1 ) + θ2 cos (ω2 t + φ2 ) , α = c1 θ1 cos (ω1t + φ1 ) + c2 θ2 cos (ω2 t + φ2 ) . Using the initial conditions, determine the values of θ 1 , φ 1 , θ 2 , and φ 2 .
Figure 1:
Cylinder Collision A hollow cylinder with mass M and radius R is at rest on a horizontal plane. In the interior of this cylinder, there is a solid disk with mass m and radius r. Initially, the center of the disk is at a distance l from the center of the cylinder and moves with velocity v ˆ y as shown in Fig. 1. Unless otherwise specified, all collisions are elastic and frictions can be ignored.
Figure 1:
1. Determine the velocity (the x ˆ and the yˆ components of the velocity) of the disk and the cylinder immediately after the first collision. Write your answer in terms of m, M , v and θ. 2. Determine the velocity (the x ˆ and the yˆ components of the velocity) of the disk and the cylinder immediately after the second collision. Write your answer in terms of m, M , v and θ. 3. If initially the disk is placed at l = (R − r)/2, determine the velocity of the disk and the cylinder immediately after the n-th collision. 4. What is the condition for l such that immediately after the n-th collision m moves with velocity v ˆ y and M is at rest? Determine the distance between two successive positions of the center of M when it is at rest. 5. For this part, the friction between the the disk and the cylinder cannot be ignored. As in part (a), initially the cylinder is at rest, while the center of the disk is at a distance l < (R − r) from the center of the cylinder and moves with velocity v ˆ y as shown in Fig. 1. If during the collision process the point of contact does not slide, determine the angular velocity of the disk and the cylinder immediately after the first collision.
Charged Discs Two thin metal discs of radius 5 cm each are suspended by electrically insulating threads such that the discs are parallel (see Fig. 1a) and close to each other (for example their distance could be 2 mm). 1. Calculate the force between the two discs if they are charged with small charges + q and − q respectively. As q is small, the displacement of the discs and the possibility of electric discharge can be neglected. 2. Now consider only one disc; calculate the surface charge distribution on a metal disc of radius R having total charge + q . (This charge distribution might be useful to answer the next question.) After this, the two original discs are each charged + q . A third metal disc of radius R > 5 cm is carefully inserted between the two discs; the third disc is neutral and is suspended by an electrically insulating thread. The three discs are all parallel to each other and their centers lie along the same horizontal line (so that when viewed head-on the discs are concentric circles). The resulting set-up is shown in Fig. 1c. ∗
3. Find the radius R of the third disc such that the net electrostatic force acting on each charged disc is zero. (The fringing effect is neglected in this problem). ∗
Figure 1: Charged discs set-up
