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A particle's position in centimeters is given by the expression x = 40 cos(45 t + 8), where t is given in seconds.

a) What is the angular frequency of this motion?

b) What is the period of this motion?

c) What is the amplitude of this motion?

d) Where is the particle at t = 0.05 s?

e) What is the velocity of the particle at t = 0.05 s?

f) What is the acceleration of the particle at t = 0.05 s?

g) What is the maximum speed attained by the particle?

Explanation / Answer

x = 40 cos(45 t + 8) compare with standard equation x = Acos(wt+Q) A=amplitude w=angular frequency Q=phase shift a)angular frequency = w = 45 b)w = 2*pi/T = 45 T = 2*pi/45 =0.139sec c)amplitude A= 40 d)at t=0.05s x = 40 cos(45 t + 8), x = 40 cos(45 *0.05 + 8) = 39.36m e)velocity = dx/dt =d/dt ( 40 cos(45 t + 8)) = -40 sin(45t+8)*45 at t=0.05 = -40*45sin(45 *0.05 +8) =-320.29m/s (-ve sign indictes retardation ) f)acceleration = d/dt (velocity) = -40*45*45 cos(45*0.05+8) = -79707.296m/sec^2

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