23 (t ) 99. a. Physical potential: = ⎣ ⎦ P R 97. a. ω = 2π ; 2 (V (2πft sin Emotional potential: ))2 ω = 2π = π ; = 0 R 28 14 (2πft ) sin V 2 2 = ω = Intellectual potential: 2π 0 R 33 b. Graphing: ⎞ V 0 2 (2πft ) = sin 0 0 2 R b. The graph is the reflected cosine graph translated up a distance equivalent to the ω amplitude. The period is 1 , so = 4π . f 2 f 1 V V 2 2 The amplitude is ⋅ = . 0 0 110 #1, #2, #3 ⎛ 2π 2 2R 50 sin ⎜ R ⎝ 23 The equation is: P (t ) ⎟ + 50 t #1 : = ⎠ (t ) P (4πft ) cos + #2 : (t ) V0 V0 = 50 sin ⎛ π ⎞ + 50 P t =− ⎜ ⎟ 2R 2R ⎝ ⎠ V 2 (4πft = ⎡1 − cos 0 )⎤ #3 : (t ) ⎞ + 50 t = 50sin ⎛ 2R P ⎠ c. No. c. Comparing the formulas: 2π (2πft ) ⎝ d. 33 sin = 1 ⎡⎣1 − cos 2 (4πft ) 110 2 #2 #1 98. a. Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus, #3 (28) = = = 2π 2 T 7305 7335 ω π 56 or = . 0 ω 28 The tunnel has a maximum height of 15 feet
97. a. 2 P (t ) = ⎣ ⎦ R 99. a. Physical potential: ω = 2π ; 23 (V sin (2πft ))2 Emotional potential: ω = 2π = π ; = 0 R V 2 sin 2 (2πft ) = 0 R Intellectual potential: b. Graphing: 28 14 ω = 2π 33 V 2 = 0 sin 2 (2πft ) R b. The graph is the reflected cosine graph translated up a distance equivalent to the amplitude. The period is 1 , so ω = 4π f . 2 f 1 V 2 V 2 The amplitude is ⋅ 0 = 0 . 110 0 0 #1, #2, #3 ⎛ 2π ⎞ 2 R 2R The equation is: #1 : P (t ) = 50 sin ⎜ ⎝ 23 t ⎟ + 50 ⎠ P (t ) V0 cos (4πft ) + V0 #2 : P (t ) = 50 sin ⎛ π t ⎞ + 50 =− ⎜ ⎟ 2R 2R ⎝ ⎠ V 2 = 0 ⎡1 − cos (4πft )⎤ 2R c. Comparing the formulas: sin 2 (2πft ) = 1 ⎡⎣1 − cos (4πft ) #3 : c. No. d. P (t ) = 50sin ⎛ 2π ⎝ 33 110 t ⎞ + 50 ⎠ 2 98. a. Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus, #2 #1 #3 T = 2π = 2 (28) = 56 ω or ω = π . 28 7305 0 7335 The tunnel has a maximum height of 15 feet
62514-0132318806_ISM_02-1.docx