Two particles of same time period (T) and amplitude undergo SHM along the same line with initial phase of $\frac{π}{6}$ . lf they start at the same instant and at same point along opposite directions. Find the time after which they will meet again for the first time:

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$\frac{T}{8}$

$\frac{T}{2}$

$\frac{T}{4}$

answer is C.

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Detailed Solution

$x = Asin (ωt + ϕ)$

$ϕ = \frac{π}{6}$

$x = Asin (ω \times 0 + \frac{π}{6}) = \frac{A}{2}$

Time taken in moving $\frac{A}{2} = \frac{T}{8}$

It is obvious that particles will meet again at Q while having covered a distance

= $4 \times \frac{A}{2} = 2 A$

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