Tuesday

September 30, 2014

September 30, 2014

Posted by **Jason** on Tuesday, November 20, 2012 at 1:41pm.

- physics -
**Damon**, Tuesday, November 20, 2012 at 2:17pmL = 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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