Question

Two particles execute shm of same amplitude of 20 cm with same period along the same line about the same equilibrium position. the maximum distance between the two is 20 cm. their phase difference in radians is

Answer 1

The phase difference between the two particles will be π/3 radians.

Explanation:

Let the wave equation of both the particles is,

x = Asin(ωt)     and

x = Asin(ωt + Ф)

where A = amplitude = 20 cm for both the particles

ω = angular velocity = same for both the particles(because period is same)

Ф = phase difference.

Distance between the two particles is given by,

d = Asin(ωt + Ф) - Asin(ωt)

=> d = A[2sin(Ф/2)cos((2ωt+Ф)/2)]

clearly we will get a maximum value when cos((2ωt+Ф)/2) = maximum = 1

hence,

d = 2A[sin(Ф/2)]

=> 20 = 2x20[sin(Ф/2)]

=> [sin(Ф/2)] = 1/2

=> Ф/2 = π/6

=> Ф = π/3

Hence the phase difference between the two particles will be π/3 radians.