SECTION 8.5 Simple Harmonic Motion; Damped Motion; Combining Waves 587 Notice that for = b 0 (zero damping), we have the formula for simple harmonic motion with amplitude a and period π ω 2 . Analyzing the Graph of an Object in Damped Motion Analyze the graph of an object in damped motion modeled by ( ) = ≥ π − d t e t t cos 0 t Solution EXAMPLE 3 The displacement d is the product of = π − y e t and = y t cos . Using properties of absolute value and the fact that ≤ t cos 1, it follows that ( ) = = ≤ = π π π π − − − − d t e t e t e e cos cos t t t t ↑ >π −e 0 t As a result, ( ) − ≤ ≤ π π − − e d t e t t This means that the graph of d lies between the graphs of = π − y e t and = − π − y e , t called the bounding curves of d. Also, the graph of d touches the graphs of the bounding curves when = t cos 1; that is, when π π = t 0, , 2 , and so on. The x-intercepts of the graph of d occur when = t cos 0; that is, at π π π 2 , 3 2 , 5 2 , and so on. See Table 1. t 0 π 2 π π3 2 π2 π −e t 1 −e 12 −e 1 −e 3 2 −e 2 t cos 1 0 −1 0 1 ( ) = π − d t e t cos t 1 0 − −e 1 0 −e 2 Point on graph of d ( ) 0, 1 π( ) 2 , 0 π( ) − −e , 1 π ( ) 3 2 , 0 π ( ) −e 2 , 2 Table 1 Figure 48 shows the graph of = y t cos , in black, the graphs of = π − y e t and = − π − y e , t the bounding curves, in red, and the graph of ( ) = π − d t e t cos t in blue. Figure 48 Damped vibration graph with bounding curves d(t) 5 e2t/p cos t y 5 cos t t 1 2p 21 p– 2 3p 2 y 5 2e2t/p y 5 e2t/p p d Exploration Graph = π − Y e x cos , x 1 along with = π − Y e x 2 and =− π − Y e , x 3 for π ≤ ≤ x 0 2 . Determine where Y1 has its first turning point (local minimum). Compare this to where Y1 intersects Y .3 Result Figure 49 shows the graphs of = = π π − − Y e x Y e cos , , x x 1 2 and =− π − Y e x 3 on a TI-84 Plus CE. Using MINIMUM, the first turning point occurs at ≈ x Y 2.83; 1 INTERSECTS Y3 at π = ≈ x 3.14. Now Work PROBLEM 23 Figure 49 Y3 5 2e2x/p Y1 5 e2x/p cosx Y2 5 e2x/p 0 21 1 2p
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