The displacement of a particle executing SHM is given by Y = 10 sin (3t +π / 3)m and ‘t’ is in seconds. The initial displacement and maximum velocity of the particle are respectively

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$5 \sqrt{3} m$ and 30 m/sec

15 m and $5 \sqrt{3} m$ /sec

$20 \sqrt{3} m$ and 30 m/sec

answer is A.

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Detailed Solution

$\begin{array}{l} y = Asin (\frac{2 π}{T}) t \Rightarrow Inition displacement (t = 0) \\ y = 10 \sin (\frac{π}{3}) = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} m \\ v_{Max} = A ω = 10 \times 3 = 30 m / \sec \end{array}$

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