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If a particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P (AP < PB) at successive intervals of 0.5 sec and 1.5 sec with a speed of 3 m/s. The maximum speed of the particle is

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a

$3 \sqrt{2} m / s$

b

6 m/s

c

3 m/s

d

$3 \sqrt{3} m / s$

answer is B.

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Detailed Solution

Time taken to move from one end to

other end $\frac{T}{2} = 2 \Rightarrow T = 4 \sec$
from mean position after 0.5sec displacement y is
$y = Asin (\frac{2 π}{4} \times 0 .5) = Asin (45^{0}) = \frac{A}{\sqrt{2}}$
velocity at $y = \frac{A}{\sqrt{2}}$
$\begin{array}{l} v = ω \sqrt{A^{2} - y^{2}}; v = ω \sqrt{A^{2} - \frac{A^{2}}{2}} = \frac{ωA}{\sqrt{2}} \\ V_{max} = ωA \Rightarrow ωA = 3 \sqrt{2} = 3 \sqrt{2} {ms}^{- 1} \end{array}$

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If a particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P (AP < PB) at successive intervals of 0.5 sec and 1.5 sec with a speed of 3 m/s. The maximum speed of the particle is