energy in shm the simple pendulum energy considerations
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Energy in SHM & The Simple Pendulum Energy Considerations in SHM - PDF document

Energy in SHM & The Simple Pendulum Energy Considerations in SHM The Simple Pendulum Homework 1 Kinetic and Potential Energy in SHM Potential energy U = 1 2 kx 2 = 1 2 kA 2 cos 2 ( t + ) U max = 1 2 kA 2 Kinetic energy


  1. Energy in SHM & The Simple Pendulum • Energy Considerations in SHM • The Simple Pendulum • Homework 1

  2. Kinetic and Potential Energy in SHM • Potential energy U = 1 2 kx 2 = 1 2 kA 2 cos 2 ( ωt + φ ) U max = 1 2 kA 2 • Kinetic energy K = 1 2 mv 2 = 1 2 mω 2 A 2 sin 2 ( ωt + φ ) K = 1 2 kA 2 sin 2 ( ωt + φ ) K max = 1 2 kA 2 2

  3. Kinetic and Potential Energy in SHM (cont’d) 3

  4. Energy Transfer in SHM 4

  5. Total Mechanical Energy in SHM E = K + U E = 1 2 kA 2 sin 2 ( ωt + φ ) + 1 2 kA 2 cos 2 ( ωt + φ ) E = 1 2 kA 2 E = K + U = 1 2 mv 2 + 1 2 kx 2 = 1 2 kA 2 � � k � � � A 2 − x 2 A 2 − x 2 � � � v = ± = ± ω � � � m 5

  6. Example A block whose mass is 0.680 kg is fastened to a spring whose spring constant is 65.0 N/m. The block is pulled a distance x = 0.110 m from its equilibrium position at x = 0 on a friction- less surface and released from rest at t = 0. (a) What is the mechanical energy of the system? (b) What are the potential energy and kinetic energy of the system when the block is at x = (1/2)x max ? 6

  7. Example A block whose mass is 0.680 kg is fastened to a spring whose spring constant is 65.0 N/m. The block is pulled a distance x = 0.110 m from its equilibrium position at x = 0 on a friction- less surface and released from rest at t = 0. (a) What is the mechanical energy of the system? (b) What are the potential energy and kinetic energy of the system when the block is at x = (1/2)x max ? 2 mv 2 + 1 (a) E = K + U = 1 2 kx 2 2 ( 65.0 N/m ) (0.110 m ) 2 = 0 . 393 J E = 0 + 1  2 = 1 2 kx 2 = 1     (b) U= 1  1  1  = 1 2 kx 2 2 k 2 x max 4 E 4 max U = 1 4 (0 . 393 J ) = 0 . 0983 J K = E − U = 0 . 393 J − 0 . 0983 J = 0 . 295 J 7

  8. The Simple Pendulum 8

  9. The Simple Pendulum (cont’d) τ = Iα − mgL sin θ = Id 2 θ dt 2 For small oscillations sin θ ≃ θ − mgLθ = mL 2 d 2 θ dt 2 d 2 θ dt 2 + g Lθ = 0 θ = θ max cos ( ωt + φ ) � g � � � ω = � � � L � T = 2 π L � � � ω = 2 π � � � g � 9

  10. Homework Set 23 - Due Mon. Nov. 8 • Read Sections 12.3 - 12.4 • Answer Questions 12.8, 12.9, 12.10 & 12.11 • Do Problems 12.15, 12.16, 12.19, 12.21, 12.22, 12.23 & 12.25 10

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