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January 31, 2015

January 31, 2015

Posted by **kylie** on Sunday, January 9, 2011 at 4:50pm.

- physics -
**drwls**, Sunday, January 9, 2011 at 7:08pmIt undergoes simple harmonic motion with an equation for angular displacement, a(t), that can be written

a(t) = A sin (2*pi*t/P)

where A is the amplitude and p is the period. I chose time zero such that the phase angle is zero at that time, to simplify the equation.

(a) w = A*(2*pi/P) cos(2*pi*t/P)

w(max) = 2*pi*(1.7 pi)/(0.46)

(b) When the displacement is half the amplitude,

2*pi*t/P = pi/6 radians

At that time, w is cos (pi/6) = (sqrt3)/2 times the amplitude

(c) When the displacement is 1/4 the amplitude, the angular argument

2*pi*t/P = 0.253 radians

The angular acceleration at any time is

alpha (t) = -A*(2*pi/P)^2*sin(2*pi*t/P)

= (A/4)(2*pi/P)^2

- physics -
**kylie**, Sunday, January 9, 2011 at 7:45pmfor part a: is it the 2*PI*(1.7PI)/.46 OR IS IT 2*pi* (1.7pi/.46)? AND FOR PART B:

DO YOU MULTIPLY THE AMPLITUDE BY (SQRT3)/2? AND I UNDERSTAND PART C

**Answer this Question**

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