- #1
applestrudle
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Homework Statement
A particle is simultaneously subjected to three SHM, all of the same frequency, and in the x direction. If the amplitudes are 0.25, 0.20, 0.15mm, respectively, and the phase difference between the 1st and 2nd is 45 degrees, and between the 2nd and 3rd is 30 degrees, find the amplitude of the resultant displacement and it's phase relative to the first (0.25mm amplitude) component.
Homework Equations
z = x + iy
z1 = x1 + iy1
x1 = 0.25 cos(wt +0)
x2 = 0.20 cos(wt +45)
x3 = 0.15 cos(wt +75)
z1 = 0.25 [x1 + i sin(wt +0)]
z2 = 0.2 [x2 + i sin(wt +45)]
z3 = 0.15 [x3 + i sin(wt +75)]
The Attempt at a Solution
Firstly, I assumed there was no initial phase angle for z1 said at t=0
x1 = 0.25
x2 = 0.20 cos(45)
x3 = 0.15 cos(75)
therefore the Amplitude must be x = x1 + x2 + x3 (as at t=0 it must be at maximum displacement so x = A)
I got x = 0.430244mm
the answers says 52mm
then i decided maybe at t=0 x is not equal to A (although I don't know why ) so i tried to calculate the magnitude of z.
z = (x^2 +y^2)^0.5
i added up all the y components at t =0 and got y = 0.286310mm
therefore z = 0.516
now for the angle:
I used arctan(y/x) = 56 degrees, the answer is ~33.5 degrees