662 CHAPTER 6 The Circular Functions and Their Graphs To develop a general equation for such motion, consider Figure 66. Suppose the point P1x, y2 moves around the circle counterclockwise at a uniform angular speed v. Assume that at time t = 0, P is at 1a, 02. The angle swept out by ray OP at time t is given by u = vt. The coordinates of point P at time t are x = a cos u = a cos vt and y = a sin u = a sin vt. As P moves around the circle from the point 1a, 02, the point Q10, y2 oscillates back and forth along the y-axis between the points 10, a2 and 10, -a2. Similarly, the point R1x, 02 oscillates back and forth between 1a, 02 and 1-a, 02. This oscillatory motion is simple harmonic motion. The amplitude of the oscillatory motion is a , and the period is 2p v . The moving points P and Q or P and R complete one oscillation, or cycle, per period. The number of cycles per unit of time, called the frequency, is the reciprocal of the period, v 2p , where v70. x y O P(x, y) (a, 0) (–a, 0) (0, a) (0, –a) R(x, 0) Q(0, y) y x u Figure 66 Simple Harmonic Motion The position of a point oscillating about an equilibrium position at time t is modeled by either s1t2 =a cos V t or s1t2 =a sin V t, where a and v are constants, with v70. The amplitude of the motion is a , the period is 2p v , and the frequency is v 2p oscillations per time unit. EXAMPLE 1 Modeling the Motion of a Spring Suppose that an object is attached to a coiled spring such as the one in Figure 65 (repeated in the margin). It is pulled down a distance of 5 in. from its equilibrium position and then released. The time for one complete oscillation is 4 sec. (a) Give an equation that models the position of the object at time t. (b) Determine the position at t = 1.5 sec. (c) Find the frequency. SOLUTION (a) When the object is released at t = 0, the distance of the object from the equilibrium position is 5 in. below equilibrium. If s1t2 is to model the motion, then s102 must equal -5. We use s1t2 = a cos vt, with a = -5. We choose the cosine function here because cos v102 = cos 0 = 1, and -5 # 1 = -5. (Had we chosen the sine function, a phase shift would have been required.) Use the fact that the period is 4 to solve for v. 2p v = 4 The period is 2p v . v = p 2 Solve for v. Thus, the motion is modeled by s1t2 = -5 cos p 2 t. a y 0 –a A. B. C. Figure 65 (repeated)
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