Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

Oscillations question

2011 · Q99

Oscillations question

2011 · Q99

View written solutionFree

Correct answer: C

Equation of SHM is given by $x = A\sin \left( {\omega t + \delta } \right)$
$\left( {\omega t + \delta } \right)$ is called phase.

When x = ${A \over 2}$ , then

$\sin \left( {\omega t + \delta } \right) = {1 \over 2}$

${ \Rightarrow \omega t + \delta = {\pi \over 6}}$

$\Rightarrow {\phi _1} = {\pi \over 6}$

For second particle,

AIPMT 2011 Mains Physics - Oscillations Question 43 English Explanation

{\phi _2} = \pi - {\pi \over 6} = {{5\pi } \over 6}

\therefore

\phi = {\phi _2} - {\phi _1}

= {{4\pi } \over 6} = {{2\pi } \over 3}

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Oscillations question

2011 · Q99

Oscillations question

2011 · Q99

View written solutionFree

Correct answer: C

Equation of SHM is given by $x = A\sin \left( {\omega t + \delta } \right)$
$\left( {\omega t + \delta } \right)$ is called phase.

When x = ${A \over 2}$ , then

$\sin \left( {\omega t + \delta } \right) = {1 \over 2}$

${ \Rightarrow \omega t + \delta = {\pi \over 6}}$

$\Rightarrow {\phi _1} = {\pi \over 6}$

For second particle,

{\phi _2} = \pi - {\pi \over 6} = {{5\pi } \over 6}

\therefore

\phi = {\phi _2} - {\phi _1}

= {{4\pi } \over 6} = {{2\pi } \over 3}