Physics 161 Lecture 17 Simple Harmonic Motion. October 30, 2018

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1 Physics 161 Lecture 17 Simple Harmonic Motion October 30,

2 Lecture 17: learning objectives Review from lecture 16 - Second law of thermodynamics. - In pv cycle process: ΔU = 0, Q add = W by gass - Q add = Q hot + Q cold - Thermal efficiency. - Entropy. (WILL NOT APPEAR IN HW OR EXAMS) This lecture 17 Kinetic Energy and Potential Energy for Hooke s Law. We will describe the relation between simple harmonic and uniform circular motions. We will find the position, velocity and acceleration of simple harmonic oscillators as a function of time. You will solve problems involving simple pendulumm. 2 Sep. 1, 2015

3 iclicker question: efficiency of heat engine What would be the thermal efficiency of a Carnot engine if the engine s hot and cold reservoirs are at the same temperature? a) 100% b) 67 % c) 50% d) 33 % e) 0 % 3 9/29/1 5

4 Hooke s Law Hooke s law: The force exerted by a spring is proportional to the negative of the displacement of the spring s end. The constant of proportionality, k, called the spring constant, depends on the spring, and has units N/m. Note that the force is NOT constant: Therefore the acceleration is NOT constant: The familiar 3 equations in kinematics are no longer true. So NO v = at NOR x = ½ a t 2 for a spring. What to do? 4 9/17/2015

5 Hooke s Law and ENERGY The spring force is a conservative force. TOTAL ENERGY IS STILL CONSERVED! Kinetic energy is still defined the familiar way, ½ m v 2. What about potential energy? Elastic potential energy: The energy associated with the spring force is given by half the product of the spring constant, k, and the displacement-squared, x /17/2015

6 Hooke s Law and ENERGY TOTAL ENERGY E = ½ m v 2 + ½ k x 2. Kinetic Energy E K and Potential Energy E P oscillate but their sum is constant. When x reaches its maximum x=a, the velocity is zero and E = ½ k A 2. When x = 0, the velocity reaches is maximum value and E = ½ m v max2. 6 9/17/2015

7 Problem: 13.3 The force constant of a spring is 137 N/m. Find the magnitude of the force required to: a) Compress the spring by 4.8 cm from its unstretched length. b) Stretch the spring by 7.36 cm from its unstretched length. 7 9/29/1 5

8 Blocks and springs A block on a spring undergoes Simple Harmonic Motion (SHM). SHM basically describes simple systems that oscillate back and forth. From conservation of energy, we can determine the speed of the block as a function of position. Simple harmonic motion: An object undergoes simple harmonic motion when it obeys Hooke s law. 8 9/17/2015

9 iclicker question: SHM energy Let the total mechanical energy of a harmonic oscillator be E 0 and the maximum displacement be A. When the displacement is A/2, what is the kinetic energy? a) E 0 /2. b) E 0 /4. c) 3E 0 /4. d) E 0. e) E 0 /8. 9 9/29/1 5

10 Describing oscillations Amplitude: A Maximum distance, x max, from equilibrium position. Period: T Time taken for one oscillation. Frequency: f Number of oscillations per second. Frequency has units of s -1 or Hertz (Hz), where 1 Hz = 1 s -1. Angular frequency: ω Number of radians per second. Angular frequency has units of rad/s /29/20

11 SHM and uniform circular motion 11 9/17/2015

12 Blocks and springs and SHM A block on a spring undergoes Simple Harmonic Motion (SHM). Period of block-spring oscillations Frequency of block-spring oscillations (f = 1/T ) Angular frequency ω of blockspring oscillations 12 9/17/2015

13 Circular motion with angular speed ω and SHM This will give the time dependence of x, v, a for SHM TIME DEPENDENCE BECAUSE v = - ω A sin ωt ω is the angular frequency given in radians per second /29/20

14 Position, velocity, and acceleration position velocity acceleration 14 9/17/2015

15 Problem: horizontal spring A horizontal object-spring system oscillates with an amplitude of 3.5 cm on a frictionless surface. If the spring constant is 250 N/m and the object has a mass of 0.5 kg, determine: a) The mechanical energy of the system. b) The maximum speed of the object. c) The maximum acceleration of the object. d) The angular frequency ω. 15 9/29/1 5

16 iclicker question: SHM(1) Which of the following is true about SHM? a) When the absolute displacement is maximum, the speed is maximum. b) When the absolute displacement is minimum, the speed is maximum. c) When the absolute displacement is maximum, the magnitude of the acceleration is minimum. d) When the magnitude of the acceleration is minimum, the speed is minimum. 16 9/29/1 5

17 iclicker question: SHM(2) An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into simple harmonic motion. What is the magnitude of the acceleration of the object when the object is at its maximum displacement of 0.10 m? a) 0 m/s 2. b) 0.45 m/s 2. c) 1.0 m/s 2. d) 2.0 m/s 2. e) 2.4 m/s /29/1 5

18 iclicker question: SHM(3) In a simple harmonic oscillator with a mass M attached to a spring with spring constant k, the mass is decreased to M/4 and the spring constant is increased to 4k. If the original period of the oscillation was T, what is the new period of oscillation? a) T. b) 4T. c) T/4. d) 16T. e) T/ /29/1 5

19 iclicker question: SHM(4) A mass of 2.0 kg is attached to a spring that has a spring constant 0f 8.0 N/m. If the mass is pulled a distance of 0.10 m and released, what is the period of the oscillation? a) 4.0 s. b) π s. c) 2π s. d) π/2 s. e) 1/(2π) s. 19 9/29/1 5

20 Simple pendulum For a small enough angles (measured in radians!), we can approximate The restoring force on the pendulum (was F = mg sinθ ) Therefore, for small angles, the pendulum approximately obeys Hooke s law. 20 9/17/2015

21 Simple pendulum and SHM A simple pendulum undergoes Simple Harmonic Motion (SHM). Period of pendulum oscillations Frequency of pendulum oscillations Angular frequency of pendulum oscillations 21 9/17/2015

22 Problem: A simple pendulum has a length of 52.0 cm and makes 82 complete oscillations in 2.0 minutes. Find: a) The period of the pendulum. b) The value of g at the location of the pendulum. 22 9/29/1 5

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