Suppose a pendulum with length L| (meters) has angle theta| (radians) from the v

Question

Suppose a pendulum with length L| (meters) has angle theta| (radians) from the vertical. It can be shown that 61 as a function of time satisfies the differential equation: d^2 theta/dt^2 + g/L sin theta = 0| where g = 9.8| m/sec/sec is the acceleration due to gravity. For small values of theta| we can use the approximation sin(theta) ~ theta|, and with that substitution, the differential equation becomes linear. Determine the equation of motion of a pendulum with length 0.5 meters and initial angle 0.5 radians and initial angular velocity d theta/dt/0.3 radians/sec. 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) seconds What is the maximum angle (in radians) from vertical? How long after reaching its maximum angle until the pendulum reaches maximum deflection in the other direction? seconds What is the period of the pendulum, that is the time for one swing back and forth?

Explanation / Answer

Ans 1.

d²/dt² + (g/L)Sin = 0 ---------- (1)
(0) = 0.5 rad, L = 0.5 m, '(0) = 0.3 rad/sec

g/L = 9.8/0.5 = 19.6

Putting in equation 1,

'' + 19.6 = 0  
m² + 19.6 = 0
m = ± i19.6
= Acos(4.43t) + Bsin(4.43t)

' = - 4.43A sin(4.43t) + 4.43B cos(4.43t)

(0) = 0.5 rad, L = 0.5 m, '(0) = 0.3 rad/sec
(0) = Acos(0) + Bsin(0) = 0.5 -> A = 0.5
'(0) = -4.43Asin(0) + 4.43Bcos(0) = 0.3

B = 0.3/4.43 = 0.06

(t) = 0.5cos(4.43t) + 0.06sin(4.43t)

b) For time to reach maximum angle,

d(t) / dt = 0

d(t) / dt = -2.22 sin 4.43 t + 0.2658 cos4.43t = 0

2.22 sin4.43 t = 0.2658 cos4.43t

tan 4.43 t = 0.12

4.43t = tan^(-1) [ 0.12]

t = 6.82/4.43 = 1.54 s

c) Maximum angle :

(t) = 0.5cos(4.43t) + 0.06sin(4.43t)
(1.54) = 0.5cos(4.43 * 1.54) + 0.06sin(4.43 * 1.54)
= 0.496 + 0.007

= 0.5 radian

d) to find d2(t) / dt

d(t) / dt = -2.22 sin 4.43 t + 0.2658 cos4.43t

d2(t) / dt = -9.83 cos 4.43t - 1.18 sin 4.43 t = 0

-9.83 cos 4.43 t = 1.18 sin 4.43 t

tan 4.43 t = -8.33

4.43 t = -83.15

t = 18.77 s but in other direction

e) T = 2pi sqrt ( L/g)

= 2 * 3.14 sqrt (0.5/9.8)

= 1.42 s

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