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A particle executes linear simple harmonic motion with an amplitude of $5 c m$ . When the particle is at $3 c m$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :-

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\frac{2}{3 π}

\frac{3 π}{2}

\frac{2 π}{3}

\frac{3}{2 π}

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detailed solution

Correct option is B

Amplitude $A = 5 cm$

When particle is at $x = 3 cm$ ,

its |velocity|=|acceleration|

i.e., $ω \sqrt{A^{2} - x^{2}} = ω^{2} x \Rightarrow ω = \frac{\sqrt{A^{2} - x^{2}}}{x}$

$T = \frac{2 π}{ω} = 2 π (\frac{3}{4}) = \frac{3 π}{2}$

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A particle executes linear simple harmonic motion with an amplitude of 5 cm. When the particle is at 3 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :-