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Q.

A particle is executing a simple harmonic motion. Its maximum acceleration is $α$ and maximum velocity is $β$ . Then its time period of vibration will be :

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a

$\frac{β^{2}}{α}$

b

$\frac{2 π β}{α}$

c

$\frac{β^{2}}{α^{2}}$

d

$\frac{α}{β}$

answer is C.

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Detailed Solution

As, we know, in SHM

Maximum acceleration of the particle, $α = A ω^{2}$

Maximum velocity. $β = A ω$

$\begin{array}{l} \Rightarrow ω = \frac{α}{β} \\ \Rightarrow T = \frac{2 π}{ω} = \frac{2 π β}{α} [ω = \frac{2 π}{T}] \end{array}$

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A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then its time period of vibration will be :