# Two particles p and q describe simple harmonic motion of same amplitude a, same frequency and along the same straight line. The maximum distance between the two particles is a√2 . The phase difference between the particle is.a)Zero b)π/2c)π/6d)π/3

Prasanna Vibhandik
45 Points
7 years ago
let the phase difference be ‘Z’
so the equation of displacement from origin for particle 1 is: y= a sin(wt+Z)
equation of displacement from origin for particle 2 is: x= a sin(wt)
So the distance between them would be y-x
a(Sin(wt+Z) – Sin(wt))
=2a[sin(Z/2)cos((2wt+Z)/2)]
this is max when the cosine part is zero
i.e 2asin(Z/2)=$\frac{a}{\sqrt{2}}$
now solve and get the ans

Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Let equation of motion to particles be given by,
x1 ​= asin(ωt) and x2 ​= asin(ωt+ϕ)
So initial phase difference between two particles is ϕ.
Now,
x2​−x1​ = a[sin(ωt+ϕ)−sin(ωt)]
= 2asin((ωt+ϕ−ωt​)/2)cos((ωt+ϕ+ωt)/2​)
= 2asin(ϕ/2)cos(ωt+ϕ/2)
Maximum value of above expression is
2asin(ϕ/2)=2​a
⟹sin(ϕ/2)=1/2
​⟹ϕ/2=π/4
⟹ϕ=π/2

Thanks and Regards