This sample exam is longer than the actual exam will be. Many of the sample exam problems are similar to those of the actual exam. = m a. I xx.

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1 Phys 326 Sample Final Exam SOLUTION 8-11 am 18 December Loomis Closed book, no calculators, but you are permitted one two-sided 8 1/2 x 11" set of notes. This sample exam is longer than the actual exam will be. Many of the sample exam problems are similar to those of the actual exam. Here are the equations you will be given on the exam. You are invited to supplement and/or annotate this list with one two-sided 8 1/2" x 11" page of your own notes. The exam is otherwise closed book/closed notes. No calculators are allowed. [K ω 2 M ]{u} = {0}; K ij = 2 x i x j V; M ij = [Z(ω)] = [K ω 2 M ]; G ij (t) = u (r) (r) sinω i u r t j r ω r u G (r) (r) i u j ij (ω) = ω 2 r ω ; 2 Im G ij (ω) = π 2ω r r 2 x i x j T [ G(ω)] = Z 1 u i (r) u j (r) δ(ω ω r ) {u (r) } T [M ]{u (s) } = δ rs [S]{u} = λ{u} {q(t)} = {u (r) }(A r cosω r t + B r sinω r t) r 1] The three degree of freedom system I xx = m a (y 2 a + z 2 a ); I xy = m a x a y a a a Γ = dl / dt = [ I ] dω / dt + ω L p α = H(q, q) = H (q, p) H (q, p) ; q α = q α p α α p α q α L(q, q) [σ ] = 3Be[I] + 2G[ε '] D Dt = t + v ρ u( r,t) = f ( r,t) + [ σ ] t = f ( r,t) + (B + G / 3) ( u) + G 2 u v + ( v ) v = g 1 p + µ ρ ρ 2 v t φ φ φ + ( p + gz) = F(t) ρ Has M and K matrices [M ] = ; [K] = It has squared natural frequencies ω 2 = , 1.746, and Find the second mode {u (2) } ( you needn't normalize it) Solution: The Eigenvalue problem for the second mode satisfies:

2 2 2 ω a 0 [K Mω 2 2 ]{u (2) 2 } = 1 3 ω 2 2 b = ω 2 c 0 We try a = 1, and conclude (from the top line) that b = (2-ω 2 2 ) = The bottom line tells us 2b + (2-2ω 2 2 )c = 0, so we conclude c = 2(2-ω 2 2 )/(2-2ω 2 2 ) = Thus {u (2) } ={ } T. The arithmetic may be checked by examining whether the middle line also sums to zero. It does.. We can also confirm that this second mode has one node, as it ought to. 2) A 2dof system is driven by a harmonic force applied to the first coordinate. [M ] d 2 x dt 2 y + [K] x y = 1 0 F o exp(iωt) The system has (normalized) modes and natural frequencies: {u (1) } = 1 1, ω 3 1 = 3; {u (2) } = 4, ω 2 = 7 Find the steady-state response of the second coordinate y(t) = Y o exp(iωt) ( You can answer this without getting the M and K matrices (getting M and K from the normalized modes and natural frequencies is possible but hard ) ( Hint: this involves [ G (ω)] ) Solution: The answer is y(t) = G 21 (ω)f o exp(iωt) so we just need to find G. If you knew M and K, you'd write G = [ K Mω 2 ] -1. You COULD get M and K from the orthonormality of the given modes, but that is not very easy. You can get G directly from its modal representation, which is G ij (ω) = u (r) (r) j u i r ω 2 r ω This is the frequency domain version of G (t) = u (r) (r) sinω 2 ij j u r t i r ω r BTW: It is this formula for G that is used to get the mode shapes and frequencies from plots of frequency-domain responses From the given mode shapes and frequencies we then construct G 21 (ω) = ω The final ω answer is given by y(t) = G 21 (ω)f o exp(iωt)

3 3) A uniform thin rod (of length L and mass M) is firmly welded at its end to a flat plate. It lies in the x-z plane of a co-rotating coordinate system. It is stuck an angle θ away from the z-axis (which is perpendicular to the plate.) The plate is spinning around the z-axis at a constant rate ω. Ignore gravity. Using the indicated xyz body coordinate system embedded in the body answer the following: a. Consider the acceleration of the rod's center of mass and determine the net force on the rod through the weld. Solution: The rod center of mass is half way up and has a distance from the axis of r= L sinθ /2. So its center of mass has an acceleration of rω 2. Thus the net force on the rod has to be Mrω 2. It points in the x direction in the picture and acts at the base. b. What are the components of the inertia tensor [ I ] for the rod around its base O? O Solution: We use the body axes x z in the picture. I xx is z 2 dm where dm = Mds/L and s is a coordinate along the bar and is integrated from 0 to L. We need to know how z depends on s: z = s cosθ. Doing the integral gives us Ixx = ML 2 cos 2 θ /3. Similarly Izz is x 2 dm where x = s sinθ, so we get Izz = ML 2 sin 2 θ /3. Ixz = - xz dm which integrates to Ixz=Izx = -ML 2 sinθ cosθ /3. We also note that Iyy is just the familiar ML 2 /3. We also note that the products of inertia Ixy and Izy are zero. c, What is the angular momentum L of the rod? Answer: {L} = [ I ] {ω} cos 2 θ 0 cosθ sinθ 0 cosθ sinθ = (ML 2 / 3) = (ML 2 ω / 3) 0 = cosθ sinθ 0 sin 2 θ ω sin 2 θ d. Consider the rate of change of the angular momentum and determine the net torque Γ applied through the weld onto the rod. In the body axes L appears not to change, so the actual change is Γ = ω x L = - j (Μω 2 L 2 /3) cosθ sinθ. This is the torque around O onto the rod through the weld. L x L y L z 4. A circular rod (length L, radius r, volumetric mass density ρ and Young's modulus E) in longitudinal vibrations is governed by the PDE ρψ (x,t) = Eψ "(x,t) Its ends are free, so it has BCs ψ '(x = L,t) = ψ '(x = 0,t) = 0

4 Find all the modes u (n) (x) and squared natural frequencies ω n 2. You need not normalize the modes. Answer: The modes satisfy ρω n 2 u (n) (x) = Eu (n) (x)" We define for convenience c = (E/ρ) 1/2 and write the ODE's solutions in the form u (n) (x) = Acosω n x / c + Bsinω n x / c The BC at the left ψ '(x = 0,t) = 0 implies that B= 0 so u (n) (x) = Acosω n x / c The BC at the rightψ '(x = L,t) = 0 then tells us 0 = (Aω n / c)sin Lω n / c If we are not to have a trivial solution then A 0 so we conclude ω n sin Lω n / c = 0. This has solutions ω n = nπc / L for any integer n. Negative n's are redundant and need not be considered ( sign can be absorbed into A) The n=0 case is admissible ( it corresponds to ω = 0 and u = constant; it is a rigid body displacement. ) The nth mode is then (unnormalized) for all n = 0,1,, u (n) (x) = cosnπ x / L ; the corresponding frequency is ω n = nπc / L. 4.' IF there were a point mass M attached to the right end of this rod, what would the boundary condition there be? Recall that strain is dψ/dx and stress is E times strain and force is stress times area. (You are not asked to find the modes for this case.) answer: Force balance on the right end tells us that M d 2 ψ / dt 2 = compression force in the rod there = -A E dψ/dx. 5. Waves in a certain peculiar unbounded 1-dimensional system are governed by the PDE αψ (x,t) = βψ "(x,t) +γ ψ "(x,t) + δ ψ '(x,t) ( α, β, γ and δ are positive constants) What is this system's dispersion relation ω = ω(k)? Answer: Substitute ψ (x,t) = exp(iωt ikx) and find: αω 2 = β k 2 +γ k 2 ω 2 + δ kω Which can be solved: ω = δk ± δ 2 k 2 + 4βk 2 (α + γ k 2 ) 2(α + γ k 2 ) Show that the system is stable, ie. that ω is real for all real k. The discriminant δ 2 k 2 + 4βk 2 (α + γ k 2 ) is positive definite; all solutions ω have zero imaginary part. We might also note, though it wasn't asked for, that this PDE is not time-reversal invariant.. because the last term changes sign when t is replaced by t; the others terms are unaffected 6. In a certain region of a solid the displacement field u is given by u x = 0.01 ( x 2 + 4xy + 2z ) u y = 0.01 ( 3x 2 2xy 2z ) u z = 0.01 ( 6x 4y ) Find the displacement gradient [ D ] ( as a function of x,y and z)

5 Answer: Here I will define D ij = u i /x j ( some people may define it with i and j interchanged; it makes no difference after we take the symmetric part ) 2x + 4y 4x 2 [ D] = x 2y 2x Find the strain tensor [ ε ] ( as a function of x,y and z) This is the symmetric part of [ D ] 2x + 4y 5x y 4 [ ε] = x y 2x Find the fractional volume change ( as a function of x,y and z) This is the trace of [ε] = 0.04 y Assuming the material is isotropic with bulk modulus B and shear modulus G, find the stress tensor [ σ ] ( as a function of x,y and z) We recall [σ ] = 3Be[I] + 2G[ε '] where e = Tr ε /3 and [ε '] = [ε] e[i]. We therefore construct 2x + 8y / 3 5x y 4 [ε '] = [ε] e[i] = x y 2x 4y / y / 3 Substitute this into [σ ] = 3Be[I] + 2G[ε '] to get 2x + 8y / 3 5x y 4 [σ ] = 3Be[I] + 2G[ε '] = 0.04By[I] G 5x y 2x 4y / y / 3 2G(2x + 8y / 3) + 4By 2G(5x y) 8G = G(5x y) 2G(2x + 4y / 3) + 4By 6G 8G 6G 8Gy / 3 + 4By Assuming the body is in static equilibrium, what must the body force density f(x,y,z) be? Answer f = [σ ]; i.e. f i = Σ j σ ji / x j f x = σ xx x + σ xy y + σ xz = 0.02G f z = σ zx z x + σ zy y + σ zz z = 0 f y = σ yx x + σ yy y + σ yz = 0.01 (22G / 3 + 4B) z 7. A plane transverse wave (i.e material displacement is perpendicular to the direction ofzpropagation) propagates in the y-direction in an unbounded homogeneous isotropic solid. The solid material has Bulk modulus B, shear modulus G and Young's modulus E. It has volumetric mass density ρ. What is the speed of that wave?

6 answer: This is a shear wave in an isotropic solid, it has speed c T = [ G/ρ] 1/2 (this follows from the equation ρu( r,t) = f ( r,t) + (B + G / 3) ( u) + G 2 u for the case that u is perpendicular to the travel direction such that. u = 0 ) An easier way to know this is just to remember. At the origin (x=y=z=0) the displacement is observed to be u(t) = k U o exp(-t 2 /τ 2 ) where τ is a constant with units of time. What is the displacement as a function of x,y,z, and t? answer:: The wave must look like u(t) = k f(t-y/c), so comparing with its known form at the origin one gets u(t) = k U o exp(-(t-y/c) 2 /τ 2 ) What is the associated stress and how does it vary in space and time? [ σ(x,y,z,t) ]? The corresponding strain is the (symmetrized) spatial derivative of this. It is only the yz and zy components of this that do not vanish: We get ε yz = (1/2) u z /y = (1/2) u z /y = U o (t-y/c)/(cτ 2 ) exp(-(t-y/c) 2 /τ 2 ) Stress is just 2G times strain. Therefore σ yz = σ zy = 2G U o (t-y/c)/(cτ 2 ) exp(-(t-y/c) 2 /τ 2 ) 8. An outgoing small amplitude spherical pressure wave in air (that has density ρ ο and wave speed c) has acoustic pressure given by p(r,t) = (A/r) cos(ω(t-r/c)) Use force balance to find the corresponding harmonic steady state material velocity v(r,t) You may need the gradient operator in spherical coordinates Solution: Force balance reads: ρ ο v/t = -p The gradient of p is p = ˆr[ ( A / r 2 )cos(ω(t r / c)) + (Aω / cr)sin(ω(t r / c)) ] Now let us divide by ρ ο and do a time integration: v = ˆr[ (A / ωρ o r 2 )sin(ω(t r / c)) + (A / ρ o cr)cos(ω(t r / c)) ] It may be noted that at large r, where the spherical wave looks locally like a plane wave, v r ~p/ρ ο c Just like in a plane wave, the ratio of pressure to velocity is ρ ο c. 9. An ideal (viscosity = 0 ) incompressible fluid of unit density ρ = 1 in 2-d has a steady timeindependent flow field v = yî + x ĵ. (The v field is pictured here.) Ignore gravity. Find how the pressure p(x,y) varies with x and y.

7 Answer: (We note, by the way, that this velocity field does indeed satisfy the claimed incompressibility v = 0.) Let us examine force balance v + v v = g 1 p + µ t ρ ρ 2 v The flow is steady so the first term drops out. There is no gravity so g drops out. There is no viscosity (the fluid is ideal), so the last term drops out. We are left with p = ρ( v ) v = ρ(v x In component form p / x = ρ{v x x + v y x v x + v y y ) v y v x}; and p / y = ρ{v x x v y + v y ( setting ρ = 1 and substituting for the components of v, this becomes p / x = x; and p / y = y These may be integrated p = x 2 / 2 y 2 / 2 + F(t) The action of F(t) is irrelevant and it can be dropped it you like, or replaced with a constant. The same answer can also be found using Bernouilli's equation in the form t φ φ φ + ( p + gz) = F(t) ρ where v = φ. y v y} The other form for Bernouilli's equation is DΨ/Dt = 0 with Ψ = 1 2 v 2 + p ρ + gz This implies merely that Ψ is constant along stream lines. So to use that argument to find pressure everywhere you have to also argue somehow that Ψ is the same for all stream lines. 10. Molasses, of density ρ and shear viscosity µ, flows in a channel formed by two parallel plates separated by a distance h. We assume the flow is incompressible, steady and laminar in the z-direction of the form v = k v(x, where 0 < x < h ). The z-axis points down, the x-axis points to the right. The channel is open to the air at the top and bottom.

8 Find the cross sectional velocity profile v(x), the pressure distribution p(x,z) and the rate Q at which volume of molasses runs out the bottom (per width into the paper.) We use the Navier-Stokes equations to describe incompressible viscid flow. We immediately observe that the assumed form v = k v(x) satisfies incompressibility v = 0 We also observe that the assumed flow is steady, so we can drop the term in v / t. We also observe that the assumed form v = k v(x) is such that the nonlinear term in Navier Stokes vanishes ( v ) v = v(x) z v(x) ˆk = 0 So force balance reduces to 0 = g ˆk 1 p + µ ˆk 2 v. The x-component of ρ ρ this tells us that p cannot depend on x; therefore p = p(y). The z component of this tells us 0 = g 1 p(z) ρ z make a constant This implies that each function must be a constant. Thus p/z = constant (which we call P'), so p = P'z + Po. and Boundary conditions on pressure at the top and bottom p = 0 tell us P' = 0. Pressure is a constant everywhere. We also have + µ v"(x) which says that a function of z p/z and a function of x, v"(x), add to ρ 0 = g 1 ρ P'+ µ ˆk v"(x) ρ Setting P' = 0, this becomes gρ = µ v"(x) So, with two constants of integration C and B, v(x) = gρ 2µ x2 + Bx + C The no-slip conditions v(x=0) = 0; v(x=h) = 0, imply C=0 and B = gρh / 2µ Thus v(x) = gρ 2µ [xh x2 ] The net flow is Q = h v(x)dx = gρ 0 12µ h3

9 11. The temperature in a certain moving fluid is found to be T(x,y,z,t) = x 2 t - y + exp(-z 2 ) The fluid has velocity v = x 2 j Show that each particle of the fluid sees a constant temperature. D Dt T = t T + v T = x 2 + x 2 ĵ {2xt î ĵ 2z exp( z2 ) ˆk} = 0 Show that this flow field is incompressible. v = y x2 = The Hamiltonian H(p x,p y,x,y) that describes the two-degree of freedom system consisting of a ramp (of mass M with horizontal Cartesian coordinate y) supporting a small cart of mass m sliding on the incline (having coordinate x corresponding to distance down the incline) takes the form H = Ap x 2 + Bp y 2 + Cp x p y + Dx ( for some constants A B C and D) Identify two constants of the motion and give expressions for them in terms of A, B, C, D, x, y, p x and p y. Because H/t = 0, we know that dh/dt = 0, so H is a constant of the motion Because H/y = 0, we know that dp y /dt = 0 and p y is a constant of the motion. 13. What does symmetry tell you about the modes of this system? Answer: The structure is not symmetric, so symmetry doesn't tell you anything.

10 14. The 1-d map x n+1 = x n 3 x n has a fixed point at x = 2 = Show whether that fixed point is stable or unstable. f(x) = x 3 x has derivative 3x 2 1. At x = 2, f' is 5. Thus f' > 1 and the point is unstable. It also has a fixed point at x = 0 Is this point stable or unstable? f(x) = x 3 x has derivative 3x 2 1. At x =0, f' is = -1. Thus f' =1 and stability is not determinable from a linear analysis near the fixed point. ( Further arguments do establish that it is indeed stable, but we never discussed how to do that in this course. ) 15. A 2dof system described by coordinates x has kinetic energy given by KE = (3/2)(dx/dt) 2 + (dx/dt)(dy/dt) + (1/2) (dy/dt) 2 What is its M matrix? Recall M ij = 2 KE so [M ] = 3 1 x i x j 1 1 Given that one of its modes is { 1 1 } T, use orthogonality to find the other. (no need to normalize it) Take the other mode in the form {u}= 1 and invoke orthogonality: a Thus a = -2 0 = {1 a} = {1 a} 4 2 = 4 + 2a