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Problem Solving Steps

Problem Solving Steps. 1. Geometry & drawing: trajectory, vectors , coordinate axes free-body diagram, … 2. Data: a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form , and

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Problem Solving Steps

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  1. Problem Solving Steps 1. Geometry & drawing:trajectory, vectors, coordinate axes free-body diagram, … 2. Data:a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!! 4. Numerical calculations and answers. 5. Check: dimensional, functional, scale, sign, … analysis of the answers and solution.

  2. Formula Sheet – PHYS 218 Mathematics π = 3.14…; 1 rad = 57.30o= 360o/2π; volume of sphere of radius R: V = (4π/3)R3 Quadratic equation: ax2 + bx + c = 0 → Vectors and trigonometry: Calculus:

  3. Chapters 1 - 3 Constants: g = 9.80 m/s2, Mearth = 6·1024 kg, c = 300 000 km/s, 1 mi = 1.6 km 1-Dimensional Kinematics: 3- or 2-Dimensional Kinematics: Equations of 1-D and 3-D kinematics for constant acceleration: Circular motion:

  4. Chapters 4 – 7 Dynamics: Energy, work, and power: Chapters 8 – 11 Momentum: Rotational kinematics:ω = dθ/dt, α = dω/dt, s = r θ, vtan = rω, atan= rα, arad= ac = rω2 Constant acceleration: ω = ω0+αt; θ = θ0+(ω0+ω)t/2, θ = θ0+ω0t+αt2/2, ω2 = ω02+2αΔθ I = Σimiri2, I=Icm+Mrcm2, KR=Iω2/2, E=Mvcm2/2+Icmω2/2+U, WR= ∫τdθ, PR=dWR/dt=τ·ω Rotational dynamics:τ = Fl= Fr sinφ, Rigid body rotating around a symmetry axis: Iαz = τz , Equilibrium: Pressure: p = F┴/A

  5. Chapters 13, 14, and 15

  6. Exam Example 1 : Coin Toss Vy=0 y (problem 2.85) Questions: (a) How high does the coin go? 0 (b) What is the total time the coin is in the air? Total time T= 2 t = 1.2 s (c) What is its velocity when it comes back at y=0 ? for y=0 and vy<0 yields vy2 = v02→ vy = -v0 = - 6m/s

  7. Exam Example 2: Accelerated Car (problems 2.7 and 2.17) Data: x(t)=αt+βt2+γt3, α=6m/s, β=1m/s2, γ = -2 m/s3, t=1s Find: (a) average and instantaneous velocities; (b) average and instantaneous accelerations; (c) a moment of time ts when the car stops. Solution: (a) v(t)=dx/dt= α+2βt +3γt2 ; v0=α; (b) a(t)=dv/dt= 2β +6γt; a0=2β; (c) v(ts)=0 → α+2βts +3γts2=0 0 x V(t) α 2β 0 t ts a(t)

  8. Exam Example 3: Truck vs. Car (problem 2.34) Data:Truck v=+20 m/s Car v0=0, ac=+3.2 m/s2 0 Questions: (a) x where car overtakes the truck; (b) velocity of the car Vc at that x; (c) x(t) graphs for both vehicles; (d) v(t) graphs for both vehicles. x Solution: truck’s position x=vt,car’s positionxc=act2/2 • x=xc when vt=act2/2 → t=2v/ac → x=2v2/ac (b) vc=v0+act → vc=2v x V(t) truck vc=2v car car truck v 0 t=2v/ac t 0 t v/ac t=2v/ac

  9. Exam Example 4: Free fall past window (problem 2.84) Data: Δt=0.42 s ↔ h=y1-y2=1.9 m, v0y=0, ay= - g y Find: (a) y1 ; (b) v1y ; (c) v2y 0 V0y=0 1st solution: (b) Eq.(3) y2=y1+v1yΔt – gΔt2/2 → v1y= -h/Δt + gΔt/2 (a)Eq.(4) → v1y2= -2gy1 → y1 = - v1y2 /2g = -h2/[2g(Δt)]2 +h/2 – g(Δt)2 /8 (c) Eq.(4) v2y2 = v1y2 +2gh = (h/Δt + gΔt/2)2 y1 ay V1y h y2 V2y 2nd solution: • Free fall time from Eq.(3): t1=(2|y1|/g)1/2 , t2=(2|y2|/g)1/2→ Δt+t1=t2 (b) Eq.(4)→ (c) Eq.(4) →

  10. Exam Example 5: Relative motion of free falling balls (problem 2.94) y 2 H Data: v0=1 m/s, H= 10 m, ay= - g Find: (a) Time of collision t; (b) Position of collision y; (c) What should be H in order v1(t)=0. 0 1 Solution: (a) Relative velocity of the balls is v0 for they have the same acceleration ay= –g → t = H/v0 (b) Eq.(3) for 2nd ball yields y = H – (1/2)gt2 = H – gH2/(2v02) (c) Eq.(1) for 1st ball yields v1 = v0 – gt = v0 – gH/v0 , hence, for v1=0 we find H = v02/g

  11. Projectile Motion ax=0 → vx=v0x=const ay= -g → voy= voy- gt x = x0 + vox t y = yo + voy t – gt2/2 v0x= v0 cos α0 v0y= v0 sin α0 tan α = vy / vx Exam Example 6: Baseball Projectile Data: v0=22m/s, α0=40o (examples 3.7-3.8, problem 3.12) Find: (a) Maximum height h; (b) Time of flight T; (c) Horizontal range R; (d) Velocity when ball hits the ground Solution: v0x=22m/s·cos40o=+17m/s; v0y=22m/s·sin40o=+14m/s • vy=0 → h = (vy2-v0y2) / (2ay)= - (14m/s)2 / (- 2 · 9.8m/s2) = +10 m • y = (v0y+vy)t / 2 → t = 2y / v0y= 2 · 10m / 14m/s = 1.45 s; T = 2t =2.9 s • R = x = v0x T = 17 m/s · 2.9 s = + 49 m • vx = v0x , vy = - v0y

  12. Exam Example 7: Ferris Wheel (problem 3.29) Data: R=14 m, v0 =3 m/s,a|| =0.5 m/s2 Find: • Centripetal acceleration • Total acceleration vector • Time of one revolution T Solution: (a) Magnitude: ac =a┴ = v2 / r Direction to center: θ (b) (c)

  13. Exam Example 8: Relative motion of a projectile and a target (problem 3.56) y Data: h=8.75 m, α=60o, vp0 =15 m/s, vtx =-0.45 m/s Find: (a) distance D to the target at the moment of shot, (b) time of flight t, (c) relative velocity at contact. Solution: relative velocity (c) Final relative velocity: (b) Time of flight (a) Initial distance x 0

  14. Exam Example 9: How to measure friction by meter and clock? d) Find also the works done on the block by friction and by gravity as well as the total work done on the block if its mass is m = 2 kg (problem 6.68).

  15. d)Work done by friction: Wf = -fkL = -μk FN L = -L μk mg cosθmax = -9 J ; work done by gravity:Wg = mgH = 10 J ; total work: W = mv||2 /2 = 2 kg (1m/s)2 /2 = 1 J = Wg + Wf = 10 J – 9 J = 1 J

  16. Exam Example 10: Blocks on the Inclines (problem 5.92) Data: m1, m2, μk, α1, α2, vx<0 m1 Find: (a) fk1x and fk2x ; (b) T1x and T2x ; (c) acceleration ax . X m2 Solution: Newton’s second law for block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1(1) block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2(2) α1 α2 X (a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v (c) T1x=-T2x, Eqs.(1)&(2)→ (b)

  17. Exam Example 11: Hoisting a Scaffold Y Data: m = 200 kg Find: (a) a force Fy to keep scaffold in rest; (b) an acceleration ay if Fy = - 400 N; (c) a length of rope in a scaffold that would allow it to go downward by 10 m 0 Solution Newton’s second law: m • Newton’s third law: Fy = - Ty , in rest ay = 0→ F(a=0)= W/5= mg/5 =392 N (b) ay= (5T-mg)/m = 5 (-Fy)/m – g = 0.2 m/s2 (c) L = 5·10 m = 50 m (pulley’s geometry)

  18. Exam Example 12: Data: L, β Find: • tension force F; • speed v; • period T. The conical pendulum(example 5.20) or a bead sliding on a vertical hoop (problem 5.119) Solution: Newton’s second law R Two equations with two unknowns: F,v Centripetal force along x: Equilibrium along y:

  19. Exam Example 13: Stopping Distance (problems 6.29, 7.29) Data: v0 = 50 mph, m = 1000 kg, μk = 0.5 Find: (a) kinetic friction force fkx ; • work done by friction W for stopping a car; • stopping distance d ; • stopping time T; • friction power P at x=0 and at x=d/2; • stopping distance d’ if v0’ = 2v0 . 0 x Solution: • Vertical equilibrium → FN = mg → friction force fkx = - μk FN = - μk mg . • Work-energy theorem → W = Kf – K0 = - (1/2)mv02 . (c) W = fkxd = - μkmgd and (b) yield μkmgd = (1/2)mv02 → d = v02 / (2μkg) . Another solution: second Newton’s law max= fkx= - μkmg → ax = - μkg and from kinematic Eq. (4) vx2=v02+2axx for vx=0 and x=d we find the same answer d = v02 / (2μkg) . (d) Kinematic Eq. (1) vx = v0 + axt yields T = - v0 /ax = v0 / μkg . • P = fkx vx → P(x=0) = -μk mgv0 and, since vx2(x=d/2) = v02-μkgd = v02 /2 , P(x=d/2) = P(x=0)/21/2 = -μkmgv0 /21/2 . (f) According to (c), d depends quadratically on v0 → d’ = (2v0)2/(2μkg) = 4d

  20. Exam Example 14: Swing(example 6.8) Find the work done by each force if (a) F supports quasi-equilibrium or • F = const , as well as the final kinetic energy K. Solution: Data: m, R, θ WT =0 always since (a) Σ Fx = 0 → F = T sinθ , Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .

  21. Exam Example 15: Riding loop-the-loop(problem 7.46) Data: R= 20 m, v0=0, m=100 kg Find: (a) min h such that a car does not fall off at point B, (b) kinetic energies for that hmin at the points B, C, and D, (c) if h = 3.5 R, compute velocity and acceleration at C. D Solution: • To avoid falling off, centripetal acceleration v2/R > g → v2 > gR. Conservation of energy: KB+2mgR=mgh → (1/2)mvB2=mg(h-2R) . Thus, 2g(h-2R) > gR → h > 5R/2 , that is hmin = 5R/2. • Kf+Uf=K0+U0 , K0=0 → KB = mghmin- 2mgR = mgR/2 , KC = mghmin- mgR = 3mgR/2 , KD = mghmin = 5mgR/2. (c) (1/2)mvC2 = KC= mgh – mgR = 2.5 mgR → vC = (5gR)1/2 ; arad = vC2/R = 5g, atan = g since the only downward force is gravity.

  22. Exam Example 16: Spring on the Incline (Fig.7.25, p.231) Data: m = 2 kg, θ = 53.1o, y0 = 4 m, k = 120 N/m, μk = 0.2, v0 =0. Find: (a) kinetic energy and speed at the 1st and 2nd passages of y=0, • the lowest position ys and friction energy losses on a way to ys, (c) the highest position yf after rebound. y m Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel y0 • 1st passage: Wnc= -y0μkmg cosθ since fk=μkFN= =μkmg cosθ, ΔK=K1 , ΔUgrav= - mgy0 sinθ, ΔUel=0 → K1=mgy0(sinθ-μkcosθ), v1=(2K1/m)1/2 =[2gy0(sinθ–μkcosθ)]1/2 2nd passage: Wnc= - (y0+2|ys|) μkmg cosθ, ΔK=K2, ΔUgrav= -mgy0sinθ,ΔUel=0 → K2=mgy0sinθ-(y0+2|ys|) μkmgcosθ, v2=(2K2/m)1/2 yf 0 ys θ (b)(1/2)kys2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) → αys2 +ys –y0 =0, where α=k/[2mg(sinθ-μkcosθ)],→ ys =[-1 - (1+4αy0)1/2]/(2α) Wnc = - (y0+|ys|) mgμkcosθ (c)Kf =0, ΔUel=0, ΔUgrav= -(y0–yf) mg sinθ, Wnc= -(y0+yf+2|ys|) μkmg cosθ →

  23. Exam Example 17: Proton Bombardment (problem 6.76) proton Data: mass m, potential energy U=α/x, initial position x0>0 and velocity v0x<0. 238U m x0 xmin 0 x Find: (a) Speed v(x) at point x. (b) How close to the repulsive uranium nucleus 238U does the proton get? (c) What is the speed of the proton when it is again at initial position x0? • Solution: Proton is repelled by 238U with a force • Newton’s 2nd law, ax=Fx/m, allows one to find trajectory x(t) as a solution • of the second order differential equation: • Easier way: conservation of energy • Turning point: v(xmin)=0 • It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)

  24. Exam Example 18: The Ballistic Pendulum(example 8.8, problem 8.43) y A block, with mass M = 1 kg, is suspended by a massless wire of length L=1m and, after completely inelastic collision with a bullet with mass m = 5 g, swings up to a maximum height y = 10 cm. L Vtop=0 Find: (a) velocity v of the block with the bullet immediately after impact; (b) tension force T immediately after impact; (c) initial velocity vx of the bullet. Solution: (a) Conservation of mechanical energy K+U=const (b) Newton’s second law yields (c) Momentum conservation for the collision

  25. Exam Example 19: Collision of Two Pendulums

  26. Exam Example 20: Head-on elastic collision (problems 8.48, 8.50) Data: m1, m2, v01x, v02x Find: (a) v1x, v2x after collision; (b) Δp1x, Δp2x , ΔK1, ΔK2 ; (c) xcm at t = 1 min after collision if at a moment of collision xcm(t=0)=0 V01x y’ V02x m1 X’ m2 Solution: In a frame of reference moving with V02x, we have V’01x= V01x- V02x, V’02x = 0, and conservations of momentum and energy yield m1V’1x+m2V’2x=m1V’01x→ V’2x=(m1/m2)(V’01x-V’1x) m1V’21x+m2V’22x=m1V’201x→ (m1/m2)(V’201x-V’21x)=V’22x= (m1/m2)2(V’01x- V’1x)2→ V’01x+V’1x=(m1/m2)(V’01x–V’1x)→ V’1x=V’01x(m1-m2)/(m1+m2) and V’2x=V’01x2m1/(m1+m2) (a) returning back to the original laboratory frame, we immediately find: V1x= V02x+(V01x– V02x)(m1-m2)/(m1+m2) and V2x = V02x +(V01x– V02x)2m1/(m1+m2) 0 X (a)Another solution: In 1-D elastic collision a relative velocity switches direction V2x-V1x=V01x-V02x. Together with momentum conservation it yields the same answer. (b) Δp1x=m1(V1x-V01x), Δp2x=m2(V2x-V02x) → Δp1x=-Δp2x(momentum conservation) ΔK1=K1-K01=(V21x-V201x)m1/2, ΔK2=K2-K02=(V22x-V202x)m2/2→ΔK1=-ΔK2(E=const) • xcm = (m1x1+m2x2)/(m1+m2) and Vcm = const = (m1V01x+m2V02x)/(m1+m2) → xcm(t) = xcm(t=0) + Vcm t = t (m1V01x+m2V02x)/(m1+m2)

  27. Exam Example 21: Head-on completely inelastic collision (problems 8.86) Data: m2=2m1, v10=v20=0, R, ignore friction Find: (a) velocity v of stuck masses immediately after collision. (b) How high above the bottom will the masses go after colliding? y Solution: (a) Momentum conservation m1 Conservation of energy: (i) for mass m1 on the way to the bottom just before the collision (ii) for the stuck together masses on the way from the bottom to the top (b) h x m2

  28. Exam Example 22: Throwing a Discus(example 9.4)

  29. Exam Example 23:Blocks descending over a massive pulley(problem 9.83) ω Data: m1, m2, μk, I, R, Δy, v0y=0 m1 R Find: (a) vy; (b) t, ay; (c) ω,α; (d) T1, T2 Solution: (a)Work-energy theorem Wnc= ΔK + ΔU, ΔU = - m2gΔy, Wnc = - μk m1g Δy , since FN1 = m1g, ΔK=K=(m1+m2)vy2/2 + Iω2/2 = (m1+m2+I/R2)vy2/2 since vy = Rω x 0 m2 ay Δy y (b) Kinematics with constant acceleration: t = 2Δy/vy , ay = vy2/(2Δy) (c) ω = vy/R , α = ay/R = vy2/(2ΔyR) (d) Newton’s second law for each block: T1x + fkx = m1ay → T1x= m1 (ay + μk g) , T2y + m2g = m2ay → T2y = - m2 (g – ay)

  30. Combined Translation and Rotation: Dynamics Note: The last equation is valid only if the axis through the center of mass is an axis of symmetry and does not change direction. Exam Example 24: Yo-Yo has Icm=MR2/2 and rolls down with ay=Rαz (examples 10.4, 10.6; problems 10.20, 10.75) Find: (a) ay, (b) vcm, (c)T Mg-T=May τz=TR=Icmαz ay=2g/3 , T=Mg/3 y ay

  31. Exam Example 25: Race of Rolling Bodies(examples 10.5, 10.7; problem 10.22, problem 10.29) y Data: Icm=cMR2, h, β Find: v, a, t, and min μs preventing from slipping FN x = h / sinβ β x Solution 1: Conservation of Energy Solution 2: Dynamics (Newton’s 2nd law) and rolling kinematics a=Rαz fs v2=2ax

  32. Equilibrium, Elasticity, and Hooke’s Law Conditions for equilibrium: Exam Example 26: Ladder against wall(example 11.3, problem 11.10) d/2 h x Static equilibrium: y State with is equilibrium but is not static. Data: m, M, d, h, y, μs Find: (a) F2, (b) F1, fs, d θ Strategy of problem solution: (c) ymanwhen ladder starts to slip (0) • Choice of the axis of rotation: arbitrary - the simpler the better. (ii) Free-body diagram: identify all external forces and their points of action. (iii) Calculate lever arm and torque for each force. (iv) Solve for unknowns. Solution: equilibrium equations yield (a) F2= Mg + mg ; (b) F1 = fs Choice of B-axis (no torque from F2 and fs) F1h = mgx + Mgd/2 → F1= g(mx+Md/2)/h = fs (c) Ladder starts to slip when μsF2 = fs, x = yd/h → μsg (M+m) = g (mymand/h+Md/2)/h →

  33. Exam Example 27: Motion in the gravitational field of two bodies (problem 13.62) M2 M1 m ΔX=X-X0 0 X X0 X2 X1 Data: masses M1, M2, m; positions x1, x2, x0, x; v(t=0)=0 Find: (a) change of the gravitational potential of the test particle m; (b) the final speed of the test particle at the final position x; (c) the acceleration of the test particle at the final position x. Solution:(a) (b) Energy conservation: (c) Newton’s 2nd law:

  34. Exam Example 28: Satellite in a Circular Orbit m Data: r = 2RE , RE = 6380 km Find: (a) derive formula for speed v and find its value; (b) derive formula for the period T and find its value; (c) satellite’s acceleration. ME r Solution: use the value g = GME/RE2 = 9.8 m/s2 RE=6380 km (a) The only centripetal force is the gravitational force: (b) The period T is a time required for one orbital revolution, that is (c) Newton’s second law with the central gravitational force yields atan = 0 and arad = ac = Fg/m = GME/r2 = (GME/RE2) (RE/r)2 = g/4 = 2.45 m/s2

  35. Exam Example 29: Satellite in an Elliptical Orbit (problem 13.77) Data: hp , ha , RE= 6380 km, ME= 6·1024 kg (apogee) (perigee) Find: (a) eccentricity of the orbit e; (b) period T; (c) arad; (d) ratio of speed at perigee to speed at apogee vp/va; (e) speed at perigee vp and speed at apogee va; (f) escape speeds at perigee v2p and at apogee v2a. 2RE hp ha Solution: (a) rp =hp+RE, ra= ha+RE, a =(rp+ra)/2, ea = a – rp, e = 1 – rp/a = 1- 2rp/(rp+ra) = = (ra- rp)/(ra+ rp) = (ha-hp)/(ha+hp+2RE) (b) Period of the elliptical orbit is the same as the period of the circular orbit with a radius equal to a semi-major axis R = a, i.e., (c) Newton’s 2nd law and law of gravitation: arad= Fgrav/ m = GME/r2. (d) Conservation of angular momentum (La= Lp) or Kepler’s second law: rava= rpvp, vp/va= ra/rp (e) Conservation of mechanical energy K + U = const : (f) Conservation of mechanical energy for an escape from a distance r (the second space speed) :

  36. Exam Example 30: A Ball Oscillating on a Vertical Spring(problems 14.38, 14.83) y Data: m, v0 , k y2=y0+A Find:(a) equilibrium position y0; (b) velocity vy when the ball is at y0; • amplitude of oscillations A; (d) angular frequency ω and period T of oscillations. Unstrained→ 0 v0 y0 Equilibrium Solution: Fy = - ky • Equation of equilibrium: Fy – mg = 0, -ky0 = mg , y0 = - mg/k (b) Conservation of total mechanical energy Lowest position y1=y0-A v1=0 (c) At the extreme positions y1,2 = y0 ± A velocity is zero and (d)

  37. Applications of the Theory of Harmonic Oscillations Oscillations of Balance Wheel in a Mechanical Watch (mass m) Newton’s 2nd law for rotation yields R Exam Example 31: SHM of a thin-rim balance wheel(problems 14.41,14.97) Data: mass m, radius R , period T Questions: a) Derive oscillator equation for a small angular displacement θ from equilibrium position starting from Newton’s 2nd law for rotation. (See above.) b) Find the moment of inertia of the balance wheel about its shaft. ( I=mR2 ) c) Find the torsion constant of the coil spring.

  38. Exam Example 32: Physical Pendulum (problem 14.99, 14.54) 0 X Data:Two identical, thin rods, each of mass m and length L, are joined at right angle to form an L-shaped object. This object is balanced on top of a sharp edge and oscillates. m m θ d Find:(a) moment of inertia for each of rods; (b) equilibrium position of the object’s center of mass; (c) derive harmonic oscillator equation for small deflection angle starting from Newton’s 2nd law for rotation; (d) angular frequency and period of oscillations. cm y Solution: (a) dm = m dx/L , (b) geometry and definition xcm=(m1x1+m2x2)/(m1+m2)→ ycm= d= 2-3/2 L, xcm=0 (c) Iαz = τz , τz = - 2mg d sinθ ≈ - 2mgd θ (d) Object’s moment of inertia

  39. Exam Example 33: Sound Intensity and Delay A rocket travels straight up with ay=const to a height r1 and produces a pulse of sound. A ground-based monitoring station measures a sound intensity I1. Later, at a height r2, the rocket produces the same second pulse of sound, an intensity of which measured by the monitoring station is I2. Find r2, velocities v1y and v2y of the rocket at the heights r1 and r2, respectively, as well as the time Δt elapsed between the two measurements. (See related problem 15.25.)

  40. Exam Example 34: Wave Equation and Transverse Waves on a Stretched String(problems 15.51 – 15.53) Data: λ, linear mass density μ, tension force F, and length L of a string 0<x<L. Questions: (a) derive the wave equation from the Newton’s 2nd law; (b) write and plot y-x graph of a wave function y(x,t) for a sinusoidal wave traveling in –x direction with an amplitude A and wavelength λ if y(x=x0, t=t0) = A; (c) find a wave number k and a wave speed v; (d) find a wave period T and an angular frequency ω; (e) find an average wave power Pav . y A L 0 X Solution: (b) y(x,t) = A cos[2π(x-x0)/λ + 2π(t-t0)/T] where T is found in (d); (c) k = 2π / λ , v = (F/μ)1/2 as is derived in (a); • v = λ / T = ω/k → T = λ /v , ω = 2π / T = kv • P(x,t) = Fyvy = - F (∂y/∂x) (∂y/∂t) = (F/v) vy2 Pav = Fω2A2 /(2v) =(1/2)(μF)1/2ω2A2. (a) Derivation of the wave equation: y(x,t) is a transverse displacement. Restoring force exerted on the segment Δx of spring: F is a tension force. μ = Δm/Δx is a linear mass density (mass per unit length). Slope= F2y/F=∂ y/∂x Newton’s 2nd law:μΔx ay= Fy , ay= ∂2y/∂t2 Slope = -F1y/F=∂y/∂x

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