Type: Double Date: Simple Harmonic Motion III. Homework: Read 10.3, Do CONCEPT QUEST #(7) Do PROBLEMS #(5, 19, 28) Ch. 10
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1 Type: Double Date: Objective: Simple Harmonic Motion II Simple Harmonic Motion III Homework: Read 10.3, Do CONCEPT QUEST #(7) Do PROBLEMS #(5, 19, 28) Ch. 10
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3 AP Physics B Mr. Mirro Simple Harmonic Motion II Date: Spiders detect prey by the vibrations of their webs, cars oscillate up and down when they hit a bump, buildings and bridges sway when the wind is fierce. As we look around us, we find many other examples of objects that vibrate or oscillate A pendulum on a grandfather clock A mass on the end of a spring A guitar string, head of a drum, reed of a saxophone or tuning fork A ruler held firmly over the edge of a table gently struck etc Since most solids are elastic, most objects vibrate when given an impulse. That is, once they are distorted, their shape tends to be restored to some equilibrium configuration. The repetitive nature of distorting and reshaping follows a pattern. These patterns, often referred to as Circadian Motion exist at many levels of the natural and physical world. We can better understand and visualize many aspects of simple harmonic motion along a straight line by looking at the relationship to uniform circular motion. By analyzing the an object rotating with UCM, we can M-O-D-E-L the periodic behavior and make conclusions about the similarity wrt SHM later on v o θ v A A 2 x 2 Top View θ Note: side (v) side ( A 2 x 2 ) As object rotates with Uniform Circular Motion side (v 0 ) A v x s h a d o w -x +x Side View A Object s shadow appears to move back and forth with Simple Harmonic Motion By similar triangles we can develop an expression relating the tangential speed to the speed of the horizontal projection. v = A 2 x 2 v o A So that v = v o A 2 x 2 = v o 1 - x 2 [ Eq. v ] where, as [x A] A A then, [v v o ]
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5 AP Physics B Mr. Mirro Simple Harmonic Motion II Date: Ex 1: Consider the displacement equation x = 0.20 cos π t [Serway13.7] 8 for a pendulum in simple harmonic motion. a. Determine the amplitude, frequency and period of the motion for this oscillator. b. What is the position of the object after two seconds have elapsed? c. What is the length of the arm of this pendulum? Ex 2: Consider a cart attached to two identical springs as shown. The cart has a mass of 1 kg and is displaced by 5 cm (m =??) to the right of equilibrium by someone reaching down to move with a horizontal force of 10 N. [Hecht10.8] a. What is the spring constant of the spring(s)? b. What is the resulting period of oscillation? c. Where will the cart be 0.2 sec after release?
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7 AP Physics B Mr. Mirro Simple Harmonic Motion III Date: When dealing with forces that are not constant, such as with simple harmonic oscillators it is often convenient and useful to use the ENERGY approach to solving problems. To stretch or compress a spring, work has to be done. Hence, potential energy is stored in a stretched or compressed spring as: PE s = ½ kx 2 Thus, the total mechanical energy E T of a mass-spring system is the sum of the kinetic and potential energies: E T = KE + PE s E T = ½ mv 2 + ½ kx 2 A m k Note: As long as there is no friction, the total mechanical energy (E T ) remains constant. While the mass oscillates back and forth, energy continuously changes from potential energy to kinetic energy and back again. At the extreme points, x = A and x = - A, all the energy is stored in the spring as potential energy. At these extreme points, the mass stops momentarily as it changes direction so v = 0! Therefore, E T = ½ m(0) 2 + ½ k(a) 2 ½ k A 2 Thus, the total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude. At the equilibrium point, x = 0 all the energy is kinetic: E T = ½ m(v) 2 + ½ k(0) 2 ½ mv 2 However, at the other intermediary points, the conservation of energy applies as follows: E stretch = E release ½ m(0) 2 + ½ k(a) 2 = ½ m(v) 2 + ½ k(x) 2 So, ½ k(a) 2 = ½ m(v) 2 + ½ k(x) 2
8 If we continue the previous expression for the total mechanical energy (E T ) of the mass-spring system, we can solve for the velocity (v) of the object at any position in terms of - that position (x), the spring constant (k), mass (m) and amplitude (A) as was done previously using uniform circular motion. k(a) 2 = m(v) 2 + k(x) 2 [ divide each term by (½) ] k(a) 2 - k(x) 2 = m(v) 2 [ transpose (k x 2 ) ] v 2 = ka 2 - kx 2 [ divide each term by (m) ] m m v 2 = k [A 2 - x 2 ] k A x 2 m m A 2 Star this *** We ll get back to it! Recalling that the maximum velocity (v max = v 0 ) occurs at the instant that the object passes through the equilibrium position (Δx = 0), we can express the gcf factor KA 2 in terms of v 0 by examining the conservation of energy. m 0 0 ½ m v ½ k Δx 2 = ½ m v min 2 + ½ k (A) 2 max min min max (½) m v 0 2 m v 0 2 = (½) k A 2 = k A 2 v 0 2 Since v 2 = k A x 2 *** m A 2 = k A 2 By substitution of v 0 m Therefore, v = v x 2 A 2 Which looks just like Eq. (v) We have since come full circle relating both UCM and SHM to the same expression for velocity for a particular position (x). On a final note The word harmonic refers to the motion being sinusoidal. The motion is simple when there is pure sinusoidal motion of a single frequency such as one tuning fork, rather than a mixture of frequencies such as our voice.
9 AP Physics B Mr. Mirro Simple Harmonic Motion III Date: Ex 1: The cone of a loudspeaker vibrates in SHM at a frequency of 262 Hz (middle C ). The amplitude at the center of the cone is A = 1.5 x 10-4 m, and at t = 0, x = A. [Giancolli11.7] a. What is equation describing the motion of the center of this cone in terms of time (t)? b. What is the maximum velocity at the center of the cone? x = 0 -x +x c. What is the maximum acceleration at the center of the cone? A d. What is the position of the cone at t = 1 millisecond 1 x 10-3 s?
10 Ex 2: A spring stretches 0.15 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.10 m from its equilibrium point and released. [Giancoli11.3] a. Determine the spring constant of the spring. b. What is the amplitude of oscillation? Equilibrium c. Using an energy approach, develop an expression for the maximum velocity v 0 of the mass in terms of amplitude (A), spring constant (k) and mass (m). d. Compute the maximum velocity for the mass in m/s. e. What is the magnitude of the velocity, v, when the mass is m from equilibrium? f. Compute the maximum acceleration of the mass. g. What is the total mechanical energy of the mass-spring system in joules?
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