The balance wheel of a watch oscillates with angular amplitude 1.7π rad and period 0.64 s. Find (a) the maximum angular speed of the wheel, (b) the angular speed of the wheel at displacement 1.7π/2 rad, and (c) the magnitude of the angular acceleration at displacement 1.7π/4 rad.

Answer 1

It 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

Answer 2

for 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