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

 

Dear Student,
Please find below the solution to your problem.

Phase difference = -2π/3

For particle 1
x1 = a.sin(wt) ...(1)
For particle 2
x2 = a.sin(wt+y) ...(2)

For instance, x1=x2=a/2

From (1)
a/2 = a.sin(wt)
Solving we get,
wt=π/6 or wt=5π/6

Putting these value in (2),
sin(π/6) = a.sin(5π/6+y) = a/2
Solving this,
y=-2π/3

Phase difference between 2 particles is -2π/3. Negative sign indicates particles moving in opposite side.

Thanks and Regards