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p2020chap07 - PHYS-2020 General Physics II Course Lecture...

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PHYS-2020: General Physics II Course Lecture Notes Section VII Dr. Donald G. Luttermoser East Tennessee State University Edition 3.3
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Abstract These class notes are designed for use of the instructor and students of the course PHYS-2020: General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille.
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VII. Vibrations and Waves A. Hooke’s Law. 1. A mass connected to a spring will experience a force described by Hooke’s Law : F s = - kx . (VII-1) a) x displacement of the mass from the unstretched ( x = 0) position. b) k spring constant . i) Stiff springs have large k values. ii) Soft springs have small k values. c) The negative sign signifies the F exerted by a spring is always directed opposite of the displacement of the mass. d) The direction of the restoring force is such that the mass is either pulled or pushed toward the equilibrium position. x = 0 x m x F x > 0 F < 0 VII–1
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VII–2 PHYS-2020: General Physics II x = 0 x m F = 0 x = 0 x m x F x < 0 F > 0 2. The oscillatory motion set up by such a system is called simple harmonic motion (SHM). a) SHM occurs when the net force along the direction of motion is a Hooke’s Law type of force. b) SHM when F ∝ - x . c) Terms of SHM: i) Amplitude ( A ): Maximum distance traveled by an object away from its equilibrium position. ii) Period ( T ): The time it takes an object in SHM to complete one cycle of motion.
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Donald G. Luttermoser, ETSU VII–3 iii) Frequency ( f ): Number of cycles per unit of time = f = 1 /T . 3. If a spring hangs downward in a gravitational field, then Eq. (VII-1) becomes F s = - kx = - mg . (VII-2) 4. In general, the equation of motion for a spring ( i.e. , SHO – simple harmonic oscillator) is F = - kx = ma (VII-3) or a = - k m x . (VII-4) Example VII–1. Problem 13.4 (Page 466) from the Ser- way & Vuille textbook : A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The spring is now placed horizontally on a table and stretched 11 cm. (a) What force is required to stretch the spring by this amount? (b) Plot a graph of force (on the y axis) versus spring displacement from the equilibrium position along the x axis. Solution (a): Here we have | F g | = mg = 50 N and | y | = 5 . 0 × 10 - 2 m. The first thing we need to do is to calculate the spring constant using Eq. (VII-2) [and using y instead of x as the independent variable] from the vertical stretch information: F s = F g - ky = - mg k = mg y = 50 N 5 . 0 × 10 - 2 m = 1 . 0 × 10 3 N/m .
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VII–4 PHYS-2020: General Physics II Now, the magnitude of the force on the spring in the horizontal direction with x = 0.11 m is F = | F s | = kx = (1 . 0 × 10 3 N/m)(0 . 11 m) = 110 N . Solution (b): Here, we are going to plot F s versus x . Since F s = - kx , we see that F s varies linearly with respect to x with the slope of the straight line being - k = - 1 . 0 × 10 3 N/m as shown in the diagram below.
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