A particle starts its SHM on a line at initial phase of $π / 6$ .It reaches again the point of start after time t. lt crosses yet another point P on the same line at successive intervals 2t and 3t respectively. Find the amplitude of the motion, if the particle crosses the point of start at speed 2 m/s:

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$\frac{2 t}{π}$

$\frac{t}{π}$

$\frac{8 t}{π}$

$\frac{4 t}{π}$

answer is A.

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Detailed Solution

$x = A \sin (ω \times 0 + \frac{π}{6}) = A \sin \frac{π}{6} = \frac{A}{2} (t = 0 at initial phase)$

Let time t=0 is considered from point Q where displacement =A/2

$\begin{matrix} Time period T = (t_{Q A} + t_{A Q}) + (t_{Q P} + t_{P Q}) + (t_{P B} + t_{B P}) \\ = t + [(2 t - t) + (2 t - t)] + t = 4 t = 2 π / ω \end{matrix}$

$\begin{array}{l} ω = π / 2 t; x = A \sin (ω t + \frac{π}{6}) \\ \frac{d x}{d t} = A ω \cos (ω t + \frac{π}{6}) = A ω \cos (\frac{π}{2} + \frac{π}{6}) \\ 2 = A \times \frac{π}{2 t} \times \frac{1}{2} (sign not considered); \\ A = \frac{8 t}{π} \end{array}$