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Two particles executing SHM of same frequency,meet at x = +A/2,while moving in opposite direction.Phase difference between the particles is.............
Pls explain.....
The phase difference should be 120 degrees(2 pi/3).
Consider the shm equation of one particle as x1 = A sin(wt) starting from origin and that of the other particle as
x2 = A sin(wt+ @). @ will be the phase difference.
particle 1 will take a time of T/12 to reach A/2 from origin with T being the timeperiod.
subst t as T/6 in the shm eqn of x1. you will get x1=A/2 and wt = pi/6 (30 degrees).
In the equation of x2, subst x2 as A/2 and wt as pi/6, u get
1/2= sin(pi/6 + @)
write the general equation:
@= n * pi + (-1)^n* pi/6 - pi/6
If u subst n=0,2,4 and so on u get @=0 which is the phase of particle 1,
if u subst n=1,3,5...... u get @=2*pi/3 which is the phase of particle 2.
therefore the phase difference will be 2 * pi/3 i.e 120 degrees
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