Question

Two particles A and B have a phase difference of π when a sine wave passes through the region.
(a) A oscillates at half the frequency of B.
(b) A and B move in opposite directions.
(c) A and B must be separated by half of the wavelength.
(d) The displacements at A and B have equal magnitudes.

Answer 1

Options (b) & (d) are Correct

Explanation:

EXPLANATION:    

Since frequencies of particles will be the same, hence, option (a) is not true.    

• Let, displacement of the particle at A be $y = A^{o} {sin(\omega t-kx)}$ & that of particle at B be $y^o = A^o sin(\omega t-kx+\pi) = - A^o sin(\omega t-kx)$

So, $y = -y^o$

• Thus, A & B moves in reverse directions & the displacements at A and B have equal magnitudes.  

So, the options (b) and (d) are true.  

When phase difference of A and B is $\pi,$  

the separation between them maybe $(n\lambda+\frac {\lambda}{2}) = (2n+1) \frac{\lambda}{2}$, where n = 1, 2, 3,.......  

Option (c) is a special case, hence it not true.

Answer 2

Answer:

It should be b, c and d.

Frequency of the two particles would be the same as frequency depends only on the source and not on the phase difference. Therefore, option a is wrong.

Since the phase difference is pie, if you draw the phase diagram then you will find that the displacement of both the particles are the same and move in opposite direction as well. (For your reference, the phase diagram is drawn in the attachment above)

A and B will be separated by half of the wavelength because in a standing wave which is though a special case, the distance between two antinodes which have a phase difference of pie between them is given as lambda/2 provided frequency is the same......

hope you understand...