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

Option 1)
$\frac{\pi }{2}$

Option 2)
$\frac{\pi }{3}$

Option 3)
$\frac{\pi }{4}$

Option 4)
$\frac{\pi }{6}$

Answers (1)

best_answer

As we learnt in

Simple harmonic as projection of circular motion -

- wherein

$x= A\cos \omega t$

$y= A\sin wt$

$x_{1}=x_{0}cos\omega t$

$x_{2}=x_{0}cos(\omega t+\phi)$

The crossing happens when $x_{1}=\frac {x_{0}}{2}$

$\frac {x_{0}}{2}=x_{0}cos \omega t=cos\omega t=\frac{1}{2}$

$cos\omega t=cos\frac{\pi}{3}$

$\omega t=\frac{\pi}{3}$

Correct option is 2.

Option 1)

$\frac{\pi }{2}$

This is an incorrect option.

Option 2)

$\frac{\pi }{3}$

This is the corect option.

Option 3)

$\frac{\pi }{4}$

This is an incorrect option.

Option 4)

$\frac{\pi }{6}$

This is an incorrect option.

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