Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x -axis. Their mean positions are separated by distance x0(.x0A.) . If the maximum separation between them is (.x0+A.) , the phase difference between their motions is

Question

Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x -axis. Their mean positions are separated by distance x0(.x0>A.) . If the maximum separation between them is (.x0+A.) , the phase difference between their motions is (A) (π /3) (B) (π /4) (C) (π /6) (D) (π /2)

Tardigrade Admin · Accepted Answer

x 1= A sin (ω t +φ1) x 2= A sin (ω t +φ2) x 1- x 2= A [2 sin [ω t +(φ1+φ2/2)] sin [(φ1-φ2/2)]] A =2 A sin ((φ1-φ2/2)) (φ1-φ2/2)=(π/6) φ1=(π/3)

Q. Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $ω$ along the $x$ -axis. Their mean positions are separated by distance $x_{0} (x_{0} > A)$ . If the maximum separation between them is $(x_{0} + A)$ , the phase difference between their motions is

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{2}$

$x_{1} = A sin (ω t + ϕ_{1})$
$x_{2} = A sin (ω t + ϕ_{2})$
$x_{1} - x_{2} = A [2 sin [ω t + \frac{ϕ _{1} + ϕ _{2}}{2}] sin [\frac{ϕ _{1} - ϕ _{2}}{2}]]$
$A = 2 A sin (\frac{ϕ _{1} - ϕ _{2}}{2})$
$\frac{ϕ _{1} - ϕ _{2}}{2} = \frac{π}{6}$
$ϕ_{1} = \frac{π}{3}$

Q. Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x -axis. Their mean positions are separated by distance x0​(x0​>A) . If the maximum separation between them is (x0​+A) , the phase difference between their motions is

3π​

4π​

6π​

2π​