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# Physics and Astronomy: Phasor diagram, particle velocity, static equilibrium, oscillator

See attached file.

#### Solution Preview

A1)
y(t) = 5sin(wt+1) + 4sin(wt+2) + 3sin(wt+3)
=> y(t) = 5sin(wt).cos(1) +5cos(wt).sin(1) + 4sin(wt).cos(2) +4cos(wt).sin(2) + 3sin(wt).cos(3) +3cos(wt).sin(3)
=> y(t) = (5cos(1)+4cos(2)+3cos(3)).sin(wt) + (5sin(1)+4sin(2)+3sin(3)).cos(wt)
let,
(5cos(1)+4cos(2)+3cos(3)) = A.cos(phi)
and,
(5sin(1)+4sin(2)+3sin(3)) = A.sin(phi)
hence, square and add:
A^2*sin^(phi) + A^2*cos^2(phi) = (5cos(1)+4cos(2)+3cos(3))^2 + (5sin(1)+4sin(2)+3sin(3))^2
=> A^2 = 25 + 16 + 9 + 40cos(1) + 24cos(1) + 30cos(2)
=> A = sqrt(50 + 64cos(1) + 30cos(2))
=> A = sqrt(50 + 64cos(1) + 60cos^2(1) - 30)
=> A = sqrt(20 + 64cos(1) + 60cos^1(1)) --Answer
phase:
tan(phi) = sin(phi)/cos(phi)
=> phi = tan-1((5sin(1)+4sin(2)+3sin(3))/(5cos(1)+4cos(2)+3cos(3)))
=> phi = 103.16 degree (see attached file) --Answer

A2.)
y1(t) = 2A.sin(wt)
y2(t) = A.sin(wt-pi/4)
w = 2*pi => time period = 1 s
v1 = dy1/dt = 2A.w.cos(wt)
v2 = dy2/dt = A.w.cos(wt-pi/4)
for v1 = v2:
2A.w.cos(wt) = A.w.cos(wt-pi/4)
=> 2.cos(wt) = cos(wt-pi/4) = cos(wt).cos(pi/4) + sin(wt).sin(pi/4)
=> (2-1/sqrt(2)).cos(wt) - sin(wt)/sqrt(2) = 0
because,
(2-1/sqrt(2))^2 + (1/sqrt(2))^2 = 5 - 2*sqrt(2)
=>sqrt((2-1/sqrt(2))^2 + (1/sqrt(2))^2) = sqrt(5 - 2*sqrt(2))
divide by this value,
((2-1/sqrt(2))/sqrt(5 - 2*sqrt(2))).cos(wt) - sin(wt)/(sqrt(2)*sqrt(5 - 2*sqrt(2))) = 0
let,
((2-1/sqrt(2))/sqrt(5 - 2*sqrt(2))) = cos(phi)
and,
1/(sqrt(2)*sqrt(5 - 2*sqrt(2))) = sin(phi)
hence,
cos(wt+phi) = 0
=> wt+phi = 2*pi*t + phi = n*pi/2
phi = tan-1((1/(sqrt(2))/(2-1/sqrt(2)))
=> phi = tan-1(1/(2.sqrt(2)-1))= tan-1(0.5469) = 28.67 degree
=> phi = 0.16*pi radian
=> t = ((n*pi/2) - (0.16*pi))/(2*pi)
=> t =((n/2) - 0.16)/2
n= +and- 1: t ...

#### Solution Summary

With good explanations and calculations, the problems are solved.

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