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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

Two particles

P

and

Q

describe SHM of same amplitude

a

and frequency

v

along the same straight line. The maximum distance between two particle is

\sqrt{2} a

. The initial phase difference between the particles is

View answer and explanation

Correct option: B

\pi / 2

Option B (

\pi / 2

) is correct because the maximum distance between the particles is related to the phase difference by

\sqrt{2}a = a\sqrt{2(1-\cos{\phi})}

.

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. At a displacement

{y_1}

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{U_1}

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{y_2}

its potential energy is

{U_2}

. The potential energy of the particle at displacement

\left( {{y_1} + {y_2}} \right)