# Composition of 2 SHM in same direction!

elllllo :D

## Homework Statement

<q>
<Q>A particle is subjected to 2 simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the individual motions,find the phase difference between the individual motions.

## Homework Equations

>Let x1 = A1sin(wt) and x2 = A2sin(wt + a)
Then, the resultant motion is also a SHM given by : x = A sin(wt +b)
where A = [(A1)2 +(A2)2 + 2(A2)(A1) cos(a) ] 1/2

## The Attempt at a Solution

Let the amplitude of individual motions was "A" and the phase difference was "d".
So, A = [ A2 + A2 + 2(A)(A)cosd ] 1/2
=> A= A [ 2 + 2cosd]1/2
=> 2 + 2cosd = 1
=> cosd = -(1/2)
=> d = 2$$\pi$$/3 or 4$$\pi$$/3 [taking values b/w [0,2$$\pi$$]
But the answer got only 2$$\pi$$/3
Did i did wrong at some place ???
Thanks for reading (^.^)

## Answers and Replies

gneill
Mentor
Take a close look at your cosine law.

Looks correct to me.

It may help you to think of these situations in terms of "phasors"
http://en.wikipedia.org/wiki/Phasor

Draw a vector representing each wave with a length equal to the amplitude, and an angle equal to the phase. The resultant vector will give you the amplitude and phase of your composition.

Thanks to you both for replying ^.^
Hmm cosine law-- U mean " cosd = -(1/2) " .
Whats wrong abt it ?? O.O
Or u wanna say i sud solve like : d = arccos(-1/2)
and take the principal values only which lies in [0,pie] and thus, get 2(pie)/3 only.
Is that so?

gneill
Mentor
Looks correct to me.

You're right of course; I was thinking of the basic cosine law without considering where the angle "a" was coming from in this problem.