Trigonometry - Can anybody walk me through this (see details)?

A support beam is subjected to vibrations along its length, emanating from two machines situated at opposite ends of the beam. The displacement caused by the vibrations can be modelled by the following equations:
𝑥1 = 3.75 sin (100𝜋𝑡 +2𝜋/9)

𝑥2 = 4.42 sin (100𝜋𝑡 −2𝜋/5)

a. State the amplitude, phase, frequency and periodic time of each of these waves. (This, I believe I know )
b. When both machines are switched on, how many seconds does it take for each machine to
produce its maximum displacement?
c. At what time does each vibration first reach a displacement of −2𝑚𝑚

Vielen Dank!

1 Answer
Sean
Mar 24, 2018

Please see below.

Explanation:

.

#color(red)((b))#

#x_1=3.75sin(100pit+(2pi)/9)#

#x_2=4.42sin(100pit-(2pi)/5)#

#x_1=3.75(sin100pitcos((2pi)/9)+cos100pitsin((2pi)/9))#

To find maximum displacement for the first machine, we take the derivative of the displacement function and set it equal to #0#:

#dx_1/dt=3.75(100picos100pitcos((2pi)/9)-100pisin100pitsin((2pi)/9))#

#dx_1/dt=375picos(100pit+(2pi)/9)=0#

#cos(100pit+(2pi)/9)=0#

#100pit+(2pi)/9=pi/2#

#100pit=(5pi)/18#

#100t=5/18#

#t=5/1800# seconds

#x_2=4.42(sin100pitcos((2pi)/5)-cos100pitsin((2pi)/5))#

#dx_2/dt=4.42(100picos100pitcos((2pi)/5)+100pisin100pitsin((2pi)/5))#

#dx_2/dt=442picos(100pit-(2pi)/5)=0#

#cos(100pit-(2pi)/5)=0#

#100pit-(2pi)/5=pi/2#

#100pit=(9pi)/10#

#100t=9/10#

#t=9/1000# seconds

#color(red)((c))#

Assuming the distances are in units of #mm#:

#-2=3.75sin(100pit+(2pi)/9)#

#sin(100pit+(2pi)/9)=-0.5333333#

#100pit+(2pi)/9=arcsin(-0.5333333)=-0.562536+pi#

#100pit=-0.562536-(2pi)/9+pi=1.8809#

#t=1.8809/(100pi)=0.00599# seconds

#-2=4.42sin(100pit-(2pi)/5)#

#sin(100pit-(2pi)/5)=-0.45249#

#100pit-(2pi)/5=arcsin(-0.45249)=-0.46956#

#100pit=0.78708#

#t=0.78708/(100pi)=0.00251# seconds

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