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**[SOLVED] the addition of three simple harmonic motion**

## Homework Statement

A particle is simultaneously subjected to three simple harmonic motions, all of the same frequency and in the x direction. If the amplitudes are 0.25, 0.20 and 0.15 mm, respectively, and the phase difference between the first and second is 45 degrees, and between the second and third is 30 degrees, find the amplitude of the resultant displacement and its phase relative to the first (0.25 mm amplitude) component.

## Homework Equations

(the sum from i=1 to n) tan x = (the sum from i=1 to n) sin x / (the sum from i=1 to n) cos x

Ae^(jrt) = A[cos(rt) + j sin(rt)], where j^2 = -1

## The Attempt at a Solution

I was able to do the first part and confirm the answer in the back of the book:

amplitude of resultant displacement ~= 0.52 mm.

for part 2, let:

first angle = x_1

second angle = x_2

third angle = x_3

i know:

|x_1 - x_2| = 45 degrees

|x_2 - x_3| = 30 degrees

thus:

x_1 = 75 degrees + x_3

x_2 = 30 degrees + x_3

I was unsure how to find the resulting phase without more information about the angles. I tried to use the above tan x formula but could not get rid of the x_3 terms...