(1 point) Suppose a pendulum with length L (meters) has angle (radians) from the

Question

(1 point) Suppose a pendulum with length L (meters) has angle (radians) from the vertical. It can be shown that as a function of time satisfies the differential equation: where g = 9.8 m/sec/sec is the acceleration due to gravity. For small values of we can use the approximation sin() ~ , and with that substitution, the differential equation becomes linear. A. Determine the equation of motion of a pendulum with length 0.5 meters and initial angle 0.4 radians and initial angular velocity deldt 0.1 radians/sec. B. At what time does the pendulum first reach its maximum angle from vertical? (You may want to use an inverse trig function in your answer) 2.458 seconds C. What is the maximum angle (in radians) from vertical? D. How long after reaching its maximum angle until the pendulum reaches maximum deflection in the other direction? (Hint: where is the next critical point?) seconds E. What is the period of the pendulum, that is the time for one swing back and forth? seconds

Explanation / Answer

A] theta = thetamax * sin (wt+ phi) where w = sqrt(g/l)

at t=0, theta =  thetamax * sin ( phi)

[theta]' = thetamax*w * cos (wt+ phi)

at t=0, [theta]'=  thetamax*w * cos ( phi)

theta/  [theta]' = 1/w tan phi and w = sqrt(9.8/0.5)=4.427

0.4/0.1 = 1/sqrt(9.8/0.5) tan phi

tan phi = 4*sqrt(9.8/0.5) = 17.709

phi = arctan 17.709 = 1.514 rad

thetamax = theta/sin phi = 0.4/sin 1.514 = 0.40065 rad

theta = 0.40065 sin( 4.427t + 1.514)

B] sin(wt+phi) = 1, wt+phi=pi/2 or t = [pi/2 -1.514]/sqrt(9.8/0.5) = 0.0128 s

C] maximum angle =  0.40065 rad

D] time required = T/2 = 2pi/2w =pi/sqrt(9.8/0.5) = 0.71 s

E] time period T =2pi/w = 2pi/sqrt(9.8/0.5) = 1.419 s

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