Physics 41 HW Set 1 Chapter 15 Serway 7 th Edition

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1 Physics HW Set Chapter 5 Serway 7 th Edition Conceptual Questions:, 3, 5,, 6, 9 Q53 You can take φ = π, or equally well, φ = π At t= 0, the particle is at its turning point on the negative side of equilibriu, at x A = Q56 Yes An oscillator with daping can vibrate at resonance with aplitude that reains constant in tie Without daping, the aplitude would increase without liit at resonance Q5 We assue the diaeter of the bob is not very sall copared to the length of the cord supporting it As the water leaks out, the center of ass of the bob oves down, increasing the effective length of the pendulu slightly lowering its frequency As the last drops of water dribble out, the center of ass of the bob hops back up to the center of the sphere, the pendulu frequency quickly increases to its original value Probles:, 3,, 3, 8, 9, 5, 9, 3, 37,, 7, 67, 69 P5 (a) x 500 c cos t π = + At 0 6 dx π = = 00 c s sint+ dt 6 v dv (c) a (d) π = = 00 c s cos t+ dt 6 A = 500 c π = 500 c cos = 33 c 6 t=, x At t= 0, v = 500 c s At t= 0, a= 73 c s π π T = = = ω P53 x = ( 00 ) cos( 300π t+ π ) Copare this with x= Acos ( ωt+ φ) to find 3 s (a) ω = π f= 300π or f= 50 Hz T = = f 0667 s (c) A = φ = 00 π rad (d) xt ( π ) = 050 s = 00 cos75 = 83

2 P5 (a) ω k 800 N = = = 00 s so position is given by x ( t) 0500 kg = 00sin 00 c Fro this we find that v ( t) = 00cos 00 c s v ax = 00 c s ( t) a= 60sin00 c s a = 60 c s ax x t= sin when x = 600 c, t= 06 s We find 00cos 00( 06 ) v = = 30 c s 960 c s a= 60sin = (c) x Using t= sin when x = 0, t= 0 when x = 800 c, t= 03 s Therefore, Δ t= 03 s P53 The 0500 s ust elapse between one turning point the other Thus the period is 00 s π ω = = 68 s T vax = ωa = ( 68 s)( 000 ) = 068 s P58 (a) F 00 N k = = = x N k ω = = 500 rad s so f= ω = 3 Hz π (c) v ωa ax = = = s at x = 0 (d) a ω A ax = = = 00 s at x = ± A = = = 00 J (e) E ka 8 (f) v ω A x (g) = = = 33 s a = ω x= 500 = 333 s 3

3 P59 Model the oscillator as a block-spring syste Fro energy considerations, v + ω x = ω A vax = ωa v = ωa so A ω + ω x = ω A 3 Fro this we find x = A 3 x= A = ± 60 c where A = 300 c P55 Using the siple haronic otion odel: π A = rθ = 5 = g 98 s ω = = = 33 rad s L (a) vax = Aω = s= 080 s a Aω ax = = s = 57 s (c) atan = rα atan 57 s α = = = r F= a= 05 kg 57 s = 06 N 57 rad s FIG P53 More precisely, (a) gh = v h= L( cosθ ) vax = gl cosθ = 087 s Iα = glsin θ α glsin θ g = = sinθi = L L ax 5 rad s Fax = gsinθ i = sin50 = 063 N (c) P59 f= 050 Hz, d = 0350, = 0 kg T = ; f I π I T = π ; T = gd gd gd gd 0( 980)( 0350) 09 kg I= T = = = π f π π 050 s FIG P535

4 P53 (a) The parallel-axis theore: I= ICM + Md = ML + Md = M + M 3 = M ( 00 ) ( 00 ) ( 3 ) 3 I M T = π = 09 s Mgd π = Mg 00 π = 980 s For the siple pendulu s 0 s T = π = 0 s difference = = 980 s 0 s 08% FIG P An 06 kg object oscillates at the end of a vertical spring which has a spring constant of 05 0 N/ The effect of air resistance is represented by the daping coefficient b = 300 N s/ (a) Calculate the frequency of the daped oscillation By what percentage does the aplitude of the oscillation decrease in each cycle? (c) Find the tie interval that elapses while the energy of the syste drops to 500% of its initial value k 05 0 N *P537 The frequency if undaped would be ω0 = = = 0s 06 kg (a) With daping b 3 kg ω = ω0 = s s 06 kg = = 0 s ω 0 f= = = 700 Hz π π s In x= A bt 0e cos ( ωt+ φ ) π over one cycle, a tie T =, the aplitude changes fro ω b to Ae π ω for a fractional decrease of A 0 0 A A e 0 0 A 0 πb ω π 3 ( 06 0 ) 000 = e = e = = = 00% (c) The energy is proportional to the square of the aplitude, so its fractional rate of decrease is twice as fast: 3t 06 bt bt 005E0 = E0e E= ka = ka 0 e = E 0 e We specify 3t = e e + 3t 06 3t 06 = 0 = ln 0 = 300 t = 06 s

5 Daping is negligible for a 050 kg object hanging fro a light 630 N/ spring A sinusoidal force with an aplitude of 70 N drives the syste At what frequency will the force ake the object vibrate with an aplitude of 00? P5 A = F ext ( ω ω 0 ) + ( bω ) With b = 0, F F F A = = = ± ext ext ext ω ω0 ( ω ω ± ) ( ω ω0 ) 0 Thus, This yields Fext k Fext 630 N 70 N ω = ω0 ± = ± = ± A A 050 kg 050 kg 00 ω = 83 rad s or ω = 03 rad s Then, f= ω gives either f= 3 Hz or f= 06 Hz π P55 Let F represent the tension in the rod (a) At the pivot, F= M g+ M g= M g y A fraction of the rod s weight Mg L as well as the weight of the ball pulls down on point P Thus, the tension in the rod at point P is y y F= M g + M g= M g + L L Relative to the pivot, I= Irod + Iball = ML + ML = ML 3 3 pivot P L y M FIG P55 I For the physical pendulu, T = π where = M d is the distance fro gd the pivot to the center of ass of the rod ball cobination Therefore, L M + M L 3L d = = M + M 3 ML π L T = π = 3 M g 3 g L ( ) For L = 00, π 00 T = = 68 s s

6 P567 (a) When the ass is displaced a distance x fro equilibriu, spring is stretched a distance x spring is stretched a distance x By Newton s third law, we expect k x = k x we find When this is cobined with the requireent that The force on either spring is given by x = x + x, FIG P57 where a is the acceleration of the ass This is in the for k x = x k+ k kk F = x= a k+ k F = keff x= a eff ( + k ) k T = π = π k kk In this case each spring is distorted by the distance x which the ass is displaced Therefore, the restoring force is F= ( k + k )x keff = k + k so that T = π ( k + k )

Physics 41 HW Set 1 Chapter 15 Serway 8 th ( 7 th )

Physics 41 HW Set 1 Chapter 15 Serway 8 th ( 7 th ) Conceptual Q: 4 (7), 7 (), 8 (6) Physics 4 HW Set Chapter 5 Serway 8 th ( 7 th ) Q4(7) Answer (c). The equilibrium position is 5 cm below the starting point. The motion is symmetric about the equilibrium