Posted by Jason on Tuesday, November 20, 2012 at 1:41pm.
L = 5 cm = .05 m
similar to k-m problem but here deflection angle = z sin (2 pi f t)
angular speed = (2 pi f)z cos (2 pi f t)
angle z from vertical
h above bottom = .05 (1-cos z)
cos z = 1 - z^2/2 + z^3/3! etc
for small angle z, cos z = 1 - z^2/2
so
h = .05(z^2/2)
potential energy at top of swing = m g h
= (m g/2) (.05 z^2)
kinetic energy at bottom = (1/2)m v^2 = (1/2)m (2 pi f .05 z)^2
so
g = (2 p f)^2 (.05)
2 pi f = sqrt (9.81/.05)
f = 2.23 Hz part b ans
T = 1/f = .449 second part a answer
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