Question

Two charged particles of same magnitude undergo simple harmonic motion having amplitude

Am and frequency (f) along parallel line. They pass each other moving in opposite direction

each time their displacement is half of their amplitude. Sketch their phase difference​

Answer 1

Answer:

π/3.

Explanation:

x1 = xmcos ωt

x2 = xmcos ωt +  

The crossing will happen when x1 = xm/2

Substituting the value of x1 = xm/2 in x1 = xmcos ωt –  

x1 = xmcos ωt

xm/2 = xmcos ωt

Cos ωt = ½

= cos  π/3

Therefore, ωt = π/3

The same constraint will happen for x2,  where wtx = 2π/3.

To find out the phase difference ? between them, substitute π/3 for wt ion the equation wt+? = 2π/3,

wt+x = 2π/3

π/3+x = 2π/3

or, x =2π/3 - π/3

= π/3

Therefore, the phase difference between the two particles will be π/3.