Wave Motion> Two particles p and q describe simple har...

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

Saurav Patil , 7 Years ago

Grade 12

2 Answers

Prasanna Vibhandik

Last Activity: 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

Last Activity: 4 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)=2a
⟹sin(ϕ/2)=1/2
⟹ϕ/2=π/4
⟹ϕ=π/2

Thanks and Regards