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11_InstSolManual_PDF_Part10

# 11_InstSolManual_PDF_Part10 - 2 21 1.45 m 21 1 2 cos 4.5°...

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11.47. Set Up: On earth, Solve: so and Reflect: Smaller g gives a larger period. 11.48. Set Up: The period is for the time for one cycle. The angular amplitude is the maximum value of Solve: (a) From the graph, From the graph we determine that the amplitude is 6.0 degrees. (b) so (c) No. The graph is unchanged if the mass of the bob is changed while the length of the pendulum and amplitude of swing are kept constant. The period is independent of the mass of the bob. 11.49. Set Up: and Solve: (a) so (b) so (c) and T unchanged, so (d) so (e) The period does not depend on m , so 11.50. Set Up: Solve: (a) The mechanical energy lost is (b) The mechanical energy has been converted to other forms by air resistance and by dissipative forces within the spring. 11.51. Set Up: As shown in Figure 11.51, the height h above the lowest point of the swing is Figure 11.51 Solve: (a) At the maximum angle of swing, and E f 5 mgL 1 1 2 cos u f 2 5 1 2.50 kg 21 9.80 m / s 2 21 1.45 m
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Unformatted text preview: 2 21 1.45 m 21 1 2 cos 4.5° 2 5 0.110 J E i 5 mgL 1 1 2 cos u i 2 5 1 2.50 kg 21 9.80 m / s 2 21 1.45 m 21 1 2 cos 11° 2 5 0.653 J E 5 mgh . K 5 u L h 5 L 2 L cos u L cos u L 1 1 2 cos u 2 . L 2 L cos u 5 h 5 E i 2 E f 5 0.30 J. E f 5 1 2 kA f 2 5 1 2 1 2.50 3 10 2 N / m 21 0.035 m 2 2 5 0.15 J E i 5 1 2 kA i 2 5 1 2 1 2.50 3 10 2 N / m 21 0.060 m 2 2 5 0.45 J E 5 1 2 kA 2 T new 5 T . T new 5 T / " 10 g S 10 g , L new 5 10 L g S 10 g L new 5 L / 9 f S 3 f , L 5 g 1 2 p f 2 2 . T new 5 " 2 T . L S 2 L , f 5 1 2 p Å g L . T 5 2 p Å L g L 5 g 1 T 2 p 2 2 5 1 9.80 m / s 2 2 1 1.60 s 2 p 2 2 5 0.635 m T 5 2 p Å L g v 5 2 p f 5 3.93 rad / s. f 5 1 T 5 0.625 Hz. T 5 1.60 s. u . T M 5 T e Å g e g M 5 1 1.60 s 2 Å 9.80 m / s 2 3.71 m / s 2 5 2.60 s. T M 2 g M 5 T e 2 g e T 2 g 5 4 p 2 L 5 constant. T 5 2 p Å L g g e 5 9.80 m / s 2 . 11-10 Chapter 11...
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