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Chapter 14 - 14.1 Solve The frequency generated by a guitar...

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14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence 3 1 1 2.27 10 s 2.27 ms 440 Hz T f = = = × =
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14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. (a) 33 s 3.3 s oscillation 3.3 s 10 oscillations T = = = (b) 1 1 0.303 Hz 0.30 Hz 3.3 s f T = = = (c) ( ) 2 2 0.303 Hz 1.90 rad s f ω π π = = = (d) The oscillation from one side to the other is equal to 60 cm 10 cm 50 cm 0.50 m. = = Thus, the amplitude is ( ) 1 2 0.50 m 0.25 m. A = = (e) The maximum speed is ( )( ) max 2 1.90 rad s 0.25 m 0.48 m s v A A T π ω = = = =
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14.3. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as ( ) cos . x t A t ω = Solve: The glider is pulled to the right and released from rest at 0 s. t = It then oscillates with a period 2.0 s T = and a maximum speed max 40 cm s 0.40 m s. v = = (a) max max 2 2 0.40 m s and rad s 0.127 m 12.7 cm 2.0 s rad s v v A A T π π ω ω π ω π = = = = = = = = (b) The glider’s position at 0.25 s t = is ( ) ( )( ) 0.25 s 0.127 m cos rad s 0.25 s 0.090 m 9.0 cm x π = = =
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14.4. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.4. Solve: (a) The amplitude 10 cm. A = (b) The time to complete one cycle is the period, hence 2.0 s T = and 1 1 0.50 Hz 2.0 s f T = = = (c) The position of an object undergoing simple harmonic motion is ( ) ( ) 0 cos . x t A t ω φ = + 0 At 0 s, 5 cm, t x = = − thus ( ) ( ) 0 1 0 0 5 cm 10 cm cos 0 s 5 cm 1 1 2 cos cos rad or 120 10 cm 2 2 3 ω φ π φ φ = + = = − = = ± ± ° Since the oscillation is originally moving to the left, 0 120 . φ = + °
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14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.5. Solve: (a) The amplitude 20 cm. A = (b) The period 4.0 s, T = thus 1 1 0.25 Hz 4.0 s f T = = = (c) The position of an object undergoing simple harmonic motion is ( ) ( ) 0 cos . x t A t ω φ = + At 0 0 s, 10 cm. t x = = Thus, ( ) 1 1 0 0 10 cm 1 10 cm 20 cm cos cos cos rad 60 20 cm 2 3 π φ φ = = = = ± = ± ° Because the object is moving to the right at 0 s, t = it is in the lower half of the circular motion diagram and thus must have a phase constant between π and 2 π radians. Therefore, 0 rad 60 . 3 π φ = − = − °
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14.6. Visualize: The phase constant 2 3 π has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is ( ) ( ) ( ) ( ) ( ) 2 0 0 3 cos cos 2 4.0 cm cos 4 rad s rad x t A t A ft t ω φ π φ π π = + = + = + The amplitude is 4 cm A = and the period is 1 0.50 s. T f = = A phase constant 0 2 3 rad 120 φ π = = ° (second quadrant) means that x starts at 1 2 A and is moving to the left (getting more negative).
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