The phase difference between two particles executing SHM of the same amplitudes and frequency along same straight line while passing one another when going in opposite directions with equal displacement from their respective stating point is 2 π / 3. If the phase of one particle is π / 6, find the displacement at this instant, if amplitude is A .

Question

The phase difference between two particles executing SHM of the same amplitudes and frequency along same straight line while passing one another when going in opposite directions with equal displacement from their respective stating point is 2 π / 3. If the phase of one particle is π / 6, find the displacement at this instant, if amplitude is A . (A) A/3 (B) 2A/3 (C) 3A/4 (D) A/2

Tardigrade Admin · Accepted Answer

Phase difference δ=(2 π/3)=δ2-δ1=δ2 =δ2-(π/6) ⇒ δ2=(5 π/6) ; sin δ1= sin δ2=(1/2) Displacement is measured form starting point (t=0) So, x=A sin (ω t+δ1) ; A sin (0+(π/6))=(A/2)

A/3

2A/3

3A/4

A/2

Phase difference δ=32π​=δ2​−δ1​=δ2​ =δ2​−6π​⇒δ2​=65π​;sinδ1​=sinδ2​=21​ Displacement is measured form starting point (t=0) So, x=Asin(ωt+δ1​);Asin(0+6π​)=2A​

Phase difference $δ = \frac{2 π}{3} = δ_{2} - δ_{1} = δ_{2}$
$= δ_{2} - \frac{π}{6} \Rightarrow δ_{2} = \frac{5 π}{6}; sin δ_{1} = sin δ_{2} = \frac{1}{2}$
Displacement is measured form starting point $(t = 0)$
So, $x = A sin (ω t + δ_{1}); A sin (0 + \frac{π}{6}) = \frac{A}{2}$