Chapter14 - 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 T f = = = × = 1 1 2 27 10 2 27 3 440 Hz s ms . .
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14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your heart’s oscillations is thus f = = = 75 beats 60 s beats s Hz 1 25 1 25 . . The period is the inverse of the frequency, hence T f = = = 1 1 0 80 1.25 Hz s .
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14.3. 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) T = = = 33 s 10 oscillations s oscillation s 3 3 3 3 . . (b) f T = = = 1 1 0 303 3.3 s Hz . (c) ω π π = = ( ) = 2 2 1 904 f 0.303 Hz rad s . (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 A = ( ) = 1 2 0 25 0.50 m m. . (e) The maximum speed is v A T A max 1.904 rad s 0.25 m m s = = = ( ) ( ) = ω π 2 0 476 .
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14.4. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as x ( t ) = A cos ω t . Solve: The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period T = 2 0 . s and a maximum speed v max cm s m s = = 40 0 40 . . (a) v A T A v max max and s rad s m s rad s m cm = = = = = = = = ω ω π π π ω π 2 2 2 0 0 40 0 127 12 7 . . . . (b) The glider’s position at t = 0.25 s is x 0.25 s 0.127 m rad s 0.25 s m cm = ( ) ( ) ( ) [ ] = = cos . . π 0 090 9 0
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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 A = 10 cm. (b) The time to complete one cycle is the period, hence T = 2 0 . s and f T = = = 1 1 2 0 0 50 . . s Hz (c) The position of an object undergoing simple harmonic motion is x t A t ( ) = + ( ) cos ω φ 0 . At s, cm, t x = = 0 5 0 thus 5 cm cm s 5 cm 10 cm rad or 60 = ( ) ( ) + [ ] = = = = ° 10 0 1 2 1 2 3 0 0 0 1 cos cos cos ω φ φ φ π
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14.6. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.6. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus f T = = = 1 1 0 25 4.0 s Hz . (c) The position of an object undergoing simple harmonic motion is x t A t ( ) = + ( ) cos ω φ 0 . At t x = = − 0 10 0 s, cm. Thus, = ( ) = = = ± = ± ° 10 1 2 2 3 120 0 0 1 1 cm 20 cm 10 cm 20 cm rad cos cos cos φ φ π Because the object is moving to the right at t = 0 s, 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, φ π 0 2 3 120 = − = − ° rad .
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14.7. 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 x t A t A ft t ( ) = + ( ) = + ( ) = ( ) ( ) + [ ] cos cos cos ω φ π φ π π 0 0 2 3 2 4 4.0 cm rad s rad The amplitude is A = 4 cm and the period is T f = = 1 0 50 . s. A phase constant φ π 0 2 3 1 = = ° rad 20 (second quadrant) means that x starts at 1 2 A and is moving to the left (getting more negative).
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