A.3.3 Given the function governing the motion of a simple pendulum with mass 0.5 kg is 0 = 0.5 cos(0.5nt+7/2) (in rad), determine: a. the period of oscillation; d. the position and linear velocity of the bob at t = 0; Answer: Oi rad and -3.12 m/s. e. the kinetic energy when the angle is 0 = 0.2 rad Answer: 4 s. b. the length of the pendulum; Answer: 3.98 m. Answer: 2.045 J. c. the maximum angle reached; f. the total mechanical energy of the pendulum. Answer: 0.5 rad or 28.66°. Answer: 2.4 J.

Physics for Scientists and Engineers: Foundations and Connections
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Chapter16: Oscillations
Section: Chapter Questions
Problem 7PQ: The equation of motion of a simple harmonic oscillator is given by x(t) = (18.0 cm) cos (10t) (16.0...
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A.3.3
Given the function governing the motion of a simple pendulum with mass 0.5 kg is 0 = 0.5 cos(0.5nt+T/2)
(in rad), determine:
a. the period of oscillation;
d. the position and linear velocity of the bob at t = 0;
Answer: 4s.
Answer: Oi rad and -3.12 m/s.
b. the length of the pendulum;
e. the kinetic energy whe the angle is 0 = 0.2 rad
Answer: 3.98 m.
Answer: 2.045 J.
c. the maximum angle reached3;
f. the total mechanical energy of the pendulum.
Answer: 0.5 rad or 28.66°.
Answer: 2.4 J.
Transcribed Image Text:A.3.3 Given the function governing the motion of a simple pendulum with mass 0.5 kg is 0 = 0.5 cos(0.5nt+T/2) (in rad), determine: a. the period of oscillation; d. the position and linear velocity of the bob at t = 0; Answer: 4s. Answer: Oi rad and -3.12 m/s. b. the length of the pendulum; e. the kinetic energy whe the angle is 0 = 0.2 rad Answer: 3.98 m. Answer: 2.045 J. c. the maximum angle reached3; f. the total mechanical energy of the pendulum. Answer: 0.5 rad or 28.66°. Answer: 2.4 J.
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