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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

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

Grade:12

2 Answers

Prasanna Vibhandik
45 Points
4 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
one year 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

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