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## 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 x

_{1}= A

_{1}sin(wt) and x

_{2}= A

_{2}sin(wt + a)

Then, the resultant motion is also a SHM given by : x = A sin(wt +b)

where A = [(A

_{1})

^{2}+(A

_{2})

^{2}+ 2(A

_{2})(A

_{1}) 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 = [ A

^{2}+ A

^{2}+ 2(A)(A)cosd ]

^{1/2}

=> A= A [ 2 + 2cosd]

^{1/2}

=> 2 + 2cosd = 1

=> cosd = -(1/2)

=> d = 2[tex]\pi[/tex]/3 or 4[tex]\pi[/tex]/3 [taking values b/w [0,2[tex]\pi[/tex]]

But the answer got only 2[tex]\pi[/tex]/3

Did i did wrong at some place ???

Thanks for reading (^.^)