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Q.

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in second is

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a

$\frac{2 π}{\sqrt{3}}$

b

$\frac{\sqrt{5}}{π}$

c

$\frac{\sqrt{5}}{2 π}$

d

$\frac{4 π}{\sqrt{5}}$

answer is C.

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Detailed Solution

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$ω^{2} y = ω \sqrt{A^{2} - y^{2}}$

$ω = angular frequency, = \frac{2 π}{T} T = Time period Now, at x = 2 cm \Rightarrow ω^{2} \times 2 = ω \sqrt{3^{2} - 2^{2}} \Rightarrow ω = \frac{\sqrt{5}}{2} \Rightarrow \frac{2 π}{T} = \frac{\sqrt{5}}{2} \Rightarrow T = \frac{4 π}{\sqrt{5}}$

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