5 - Simple Harmonic Motion | Lecture Note

University:
High School
Course:
High School Physics
Academic year:

2024
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437

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8
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5: Pendulums, Simple Harmonic Motion, and Periodic Motions AP FIZZ Motivation • We discuss centripetal motion in terms of tangential velocity. Can we re-frame it in terms of period, and what other motions are periodic? Centripetal Motion… re-framed • We know that 𝐹𝐶 = • • • 𝑚𝑣 2 , and 𝑎𝑐 𝑟 = 𝑣2 𝑟 But, v is circumference of a circle divided by time around the same circle, or 2𝜋𝑟/𝑇, so re-state as 𝑎𝑐 = 4𝜋2 𝑟 , or 𝑇 𝑇2 = 2𝜋 𝑟 𝑎𝑐 Motions with fixed periods (undamped) are said to be simple harmonic motions. Simple Pendulum • Consider a simple (massless, small angle of motion) pendulum. Period of one cycle depends on length of pendulum arm, and local gravity. 𝑇 = 2𝜋 𝐿 𝑔 EX 1: A simple pendulum takes 10.0s to complete 4 full cycles. What is the length of the pendulum arm? 10 𝑇= = 2.50𝑠 4 𝐿 𝑇 = 2𝜋 𝑔 𝑇2 𝐿= = 4𝜋 2 𝐿 𝑔 (9.80)(2.50)2 4𝜋2 𝐿 = 1.55𝑚 Oscillating Spring (Simple Harmonic Motion) • • Consider a mass, m, on a spring with spring constant, k. If mass horizontally oscillates back and forth on frictionless surface, then Hooke’s Law force is equal to centripetal force. Label maximum amplitude of motion as A. Then 𝑇 = 2𝜋 𝑚 𝑘 And also energy is conserved such that 1 2 1 1 2 2 𝐸𝑇 = 𝑘𝐴 = 𝑚𝑣 + 𝑘𝑥 2 2 2 EX 2: A 0.45kg mass is attached to a spring with k=140N/m. It horizontally oscillates on a frictionless surface. The mass is displaced 15.0cm and is then released. Determine: a) period of motion, b) speed of mass at equilibrium position, and c) amplitude of motion when mass has a speed of 1.0m/s. 𝑚 𝑇 = 2𝜋 𝑘 0.45 𝑇 = 2𝜋 = 0.356 𝑠 140 1 2 𝐸𝑇 = 𝑘𝐴 = 0.5 140 0.15 2 = 1.575 𝐽 2 1 1 1.575 = 𝑚𝑣 2 + 𝑘(0)2 2 2 𝑣 = 2.65 𝑚/𝑠 EX 2: A 0.45kg mass is attached to a spring with k=140N/m. It horizontally oscillates on a frictionless surface. The mass is displaced 15.0cm and is then released. Determine: a) period of motion, b) speed of mass at equilibrium position, and c) amplitude of motion when mass has a speed of 1.0m/s. 1 1 2 2 1.575 = 𝑚𝑣 + 𝑘𝑥 2 2 1.575 = 0.5 0.45 1.0 2 1 2 + 𝑘𝑥 2 1 2 1.575 = 0.225 + 𝑘𝑥 2 1.35 = (0.5)(140)𝑥 2 𝑥 = 0.139 𝑚