#x_1 = 3.75 sin (50 (2pi t) + {2pi}/9)#
That goes from #-3.75# to #+3.75#, usually called an amplitude of #3.75,# the scale factor on the sine.
The phase shift is an #(-{2pi}/9 )/(50(2pi))=-1/450#, the minus sign indicating a lagging phase shift.
Assuming #t# is in seconds, the frequency is 50 Hertz.
The period is thus #1/50=.02# seconds
#x_2 = 4.42 sin (50(2pi t) - {2pi}/5) #
Amplitude #4.42#, phase shift #{ {2pi}/5 } / {50(2pi) } = 1/250 # leading, frequency and period the same as the first.
Each gets maximum displacement when the argument is #pi/2# because #sin(pi/2)=1# is the maximum of the sine.
# 50 (2pi t) + {2pi}/9 = pi/2#
#t = {pi/2 - {2pi}/9}/{100 pi} = 1/360 # seconds for machine 1
# 50 (2pi t) - {2pi}/5 = pi/2#
#t = {pi/2 + {2pi}/5}/{100 pi} = 9/1000 # seconds for machine 2
When do these reach #-2# mm? Just like with #t# we're never really told the units of #x#; let's assume mm.
#-2 = 3.75 sin (50 (2pi t) + {2pi}/9)#
# sin(50(2pi t) + {2pi}/9 = -2/ 3.75#
# t = { arcsin(-2/3.75) - {2pi}/9} / {50(2pi) } #
That's negative, so we go to the next inverse sine:
# t = { (pi-arcsin(-2/3.75)) - {2pi}/9} / {50(2pi) } approx 0.009568# sec for machine 1 to get to #-2.#
#-2 = 4.42 sin (50(2pi t) - {2pi}/5) #
#t = { arcsin(-2/4.42) + {2pi}/5 }/{50(2pi)} approx 0.002505 # seconds for machine 2 to get to #-2#.
#x_1 = 3.75 sin ( {2pi}/9 + 50 (2pi t) )#
#x_1 = 3.75 cos( {2pi}/9) sin (50 (2pi t)) + 3.75 sin ({2pi}/9) cos(50 (2pi t))#
#A=3.75 cos( {2pi}/9), quad B= 3.75 sin ({2pi}/9)#
#x_2 = 4.42 sin (- {2pi}/5 + 50(2pi t) ) #
#x_2=4.42 sin(-{2pi}/5) cos(50(2pi t)) + 4.42 cos(-{2pi}/5) sin(50((2pi t)) #
#72^circ# actually has a reasonable closed form form its trig functions, being related to the constructible pentagon. #40^circ# isn't a constructible angle, so I'm guessing we're long past when we were supposed to start approximating. I never like that part.
#x_1+x_2 = ( 3.75 cos( {2pi}/9) + 4.42 cos({2pi}/5) ) sin (50 (2pi t)) + (3.75 sin ({2pi}/9) - 4.42 sin({2pi}/5) ) cos(50 (2pi t)) #
# approx 4.23852 sin (50 (2pi t)) -1.79322 cos (50 (2pi t)) #
We're supposed to recognize this as the sine difference angle formula with a scale factor.
# r sin(a-b) = r sin a cos b - r cos a sin b = r cos b sin a - r sin b cos a #
We turn the coefficients to polar coordinates. Since we're using the difference formula we're in the first quadrant:
# r= sqrt{ 4.23852 ^ 2 + 1.79322^2 } = 4.60225 #
# theta = arctan( 1.79322/4.23852 ) = 0.400241# radians #= .1274 pi# radians
We engineered #r cos theta = 4.23852# and #r sin theta = 1.79322# so we can substitute:
# r sin(a-theta) = r cos theta sin a - r sin theta cos a #
#x_1+x_2 = 4.60225 sin(50 (2pi t) - 0.400241) #
I should check all this but I'm not going to.
