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Two particles P(1) and P(2) are performi...

Two particles `P_(1)` and `P_(2)` are performing `SHM` along the same line about the same meabn position , initial they are at their position exterm position. If the time period of each particle is `12` sec and the difference of their amplitude is `12 cm` then find the minimum time after which the seopration between the particle becomes `6 cm`

Text Solution

Verified by Experts

The correct Answer is:
`t = 2s`
The coordinates of the particeles are
`x_(1) = A_(1) cos omegat, x_(2) = A_(2) cos omegat`
seperation `= x_(1) - x_(2) = (A_(1) - A_(2)) cos omegat = 12 omegat`
Now `x_(1) - x_(2) = 6 = 12 cos omegat`
`rArr omegat = (pi)/(3)`
`rArr (2pi)/(12). t = (pi)/(3)`
`rArr t = 2s`
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