exam2_solutions

# exam2_solutions - Exam 2 for PHYSICS 2210 ...

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Unformatted text preview: Exam 2 for PHYSICS 2210  9:30 to 10:45 am on Nov 5. 2009    Name:                                                                                               Score:    1) Multiple‐choice questions (30 points total, and 5 points each question).    1a) For r, a, and b real constants, the complex conjugate of  re a +ib  is:    A)  re-a +ib ,              B)  -re a +ib ,               € -a +ib C)  re ,              a −ib D)  re ,  -a −ib E)  re ,                Answer: D)    1b) For 2nd order, inhomogeneous, ordinary differential equation:   d2x dx + 3 + 2x = 5e− t . For arbitrary constant k, which of the following form is  dt dt 2 correct for the particular solution, xp(t)?    A)  ke-t ,                 B)  ke-2t ,               C)  kte-t ,                   D)  kte-2t ,                 E)  kt 2e-t .  Answer: C)    1c) For a driven oscillator (spring‐block) with damping, there is a resonant  frequency ωR. If the driving force is at this resonant frequency, the amplitude of the  response is maximized. Suppose that the spring constant is k, the mass of the block  is m, damping parameter is b, and the magnitude of driving force is F0. Which of the  following statement is true for ωR?    A) ωR depends on k, m, b, and F0.   B) ωR depends on k, m, and b.    C) ωR depends on k, m and F0.    D) ωR depends on k and m.  E) None of above.    Answer: B)      € € € € € € € € € € €   1  x(t) or G(t,t’) 1d) The top left panel shows the response of an oscillator to an impulse force, i.e.,  the Green’s function, G(t,t’). Suppose that a force function, f(t), consists of two  impulses applied at two distinct times (t=2 and 3), shown in the top right panel.     Among A, B, C, and D, what is the correct response of the oscillator to this force f(t)?    1 1 0 Force f(t) 0 −1 1 0 −1 1 0 0 1 2 t-t’3 4 5 6 −1 1 0 0 1 2 t 3 4 5 6 A) B) x(t) 0 1 2 t3 4 5 6 x(t) −1 1 0 0 1 2 t 3 4 5 6 C) D) x(t) −1   Answer: C)       1e) How is the block’s kinetic energy KE, of an oscillator (spring‐block) dependent  on quality factor Q of the oscillator?     A)  KE decays with time faster for larger Q  B)  KE decays with time slower for larger Q  C)  KE decays with time but does not depend on Q.  D)  KE remains constant with time, independent of Q.    Answer: B)       0 1 2 t 3 4 5 6 x(t) −1 0 1 2 t 3 4 5 6   2  1f) Consider a spring‐block oscillator with no damping. Suppose that the spring  k , for x > 0 constant k is given as:  k = 0  , where k0 is a positive constant,  Which  0.5k0 , for x < 0 of the following phase diagrams is correct?    v v A) B) € x x C) v D) v x x      Answer: B)     3  2) (50 points total) This problem has four parts. A block‐spring oscillator (see figure  below) is placed in a water tank. The spring constant is k, the block’s mass is m, and  the resisting/damping force on the block is given as  Fr = −bv , where b>0 and v is  velocity. Suppose that at the beginning (i.e., t=0), the block is at the equilibrium  position but moves to the right side at a speed of v0. Suppose that the oscillation is a  under­damping system.   €   2a) (15 points) (i) Setup a coordinate system, by  k specifying and drawing the origin and positive  m direction of the coordinate system. (ii) Write down  the Newton’s second law as a second order  differential equation. (iii) What is the condition for    under‐damping?       Answer: Let x be the displacement from the  equilibrium position of the block, with positive x  pointing to the right side (I ignore the drawing  here). The Newton’s second law in this coodinate system is   b k b k ˙ ˙ ˙ m˙˙ = −kx − bx , or   ˙˙ + x + x = 0 , or  ˙˙ + 2βx + ω 2 x = 0 , where  β = x x , ω 2 = .  x 0 0 m m 2m m 2/(4m). or  β 2 < ω 2 .  For under‐damping, k>b €   € € €   2b) (15 points) Solve the equation for the solution of oscillation with time, using the  € initial conditions (Hint: Use the cosine or sine function form for the general solution.  If you introduce new variables such as β or ω, you need to state how they are related  to k, m, and b).    Answer: The solution before considering initial condition is  0 x(t) = Ae- βt cos(ωt - δ ) , where  ω = ω 2 − β 2 , and A and δ are constants to be  0 € € € determined.  At t=0, x(0)=0, and v(0)=v0.  ˙ v(t) = x(t) = -Ae- βt [ βcos(ωt - δ ) + ωsin(ωt - δ )] .  € From x(0)=0, δ=π/2. And from v(0)=v0, A=v0/ω.  v π v Therefore,  x(t) = 0 e- βt cos(ωt - ) = 0 e- βt sin(ωt) , and  ω 2 ω v0 - βt ˙ v(t) = x(t) = - e [ βsin(ωt) - ωcos(ωt)] .  ω   €           4  c) (10 points) Sketch the phase diagram.   v v0 x € €                   d) (10 points) The total mechanical energy, E, of the oscillation system decays with  time, due to damping. Derive an expression for the rate of the energy decay with  time: dE/dt, using the results from part b).    1 1 Answer:   E = T + U = mv 2 + kx 2 .  2 2 dE ˙ ˙ ˙x ˙˙x ˙ = mvv + kxx = mx˙˙ + kxx = x(m˙˙ + kx) = -bx 2 < 0 ,   dt ˙ x where we considered  m˙˙ = −kx − bx .  € 2 bv 0 dE dE ˙ = - 2 e-2 βt [ βsin(ωt) - ωcos(ωt)] 2 .  = -bx 2     dt dt ω   €      €     5  3) (20 points) A wooden block with uniform cross section area, A, and a height H,  floats in a water pool at the equilibrium position. Suppose that the densities for the  block and water are ρb and ρ0, respectively. The block is perturbed slightly by  pulling the block up a little and then releasing it. Assume that the viscous resisting  force and frictional force can be ignored. i) Set up a coordinate system (the origin  and the positive direction), and write down the Newton’s 2nd law for the block, ii)  demonstrate that the block undergoes oscillatory motion after the perturbation, and  iii) determine the period of the oscillation and express the solution in terms of the  variables given above (g is the gravitational acceleration).     air H ρb ρ0 water € € €       Answer:  Set up a coordinate system with the origin at the center of mass of the  block at the equilibrium, and positive direction pointing upward (drawing is ignored  here).   First, at the equilibrium, the weight of the block is balanced by buoyancy.   mg = Fb_e ,   where m=ρbHA, is the mass of the block, and Fb_e=ρ0heAg, is the buoyancy at the  equilibrium, where he is the height of the block under the water at equilibrium.  he=Hρb/ρ0.   After the perturbation is introduced, the Newton’s 2nd law is  m˙˙ = −mg + Fb ,   x where the buoyancy force Fb=ρ0(he‐x)Ag=ρ0heAg‐ρ0xAg=mg‐ρ0xAg. Therefore,   ρg x ρ b HA˙˙ = − ρ 0 gAx    or  ˙˙ + 0 x = 0 .  x ρb H This equation is the standard form for oscillation with an angular frequency   ρ0g 2π ρH   and period  T = ω= = 2π b .  ρb H € ω ρ0g   € €   6  ...
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