Q20 / 107 · NTA Abhyas (JEE Main) 2020 · Physics · Oscillations

Two particles P1 and P2 are performing SHM along the same line about the same mean position. Initially, they are at their positive extreme position. If the time period of each particle is 12 s and the difference of their amplitudes is 12 cm then find the minimum time (in seconds) after which the separation between the particles become 6 cm.

Correct answer: 2

Solution

The x coordinates of the particles are

x1=A1cos⁡ωt, x2=A2cos⁡ωt

Separation x1-x2=A1-A2cos⁡ωt=12cos⁡ωt

Now x1-x2=6=12cos⁡ωt

⇒ωt=π3⇒2π12 ×t=π3

⇒t=2 s

Hence, the correct answer is 2.

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