A particle is executing a simple harmonic motion. Its maximum acceleration is and maximum velocity is . Then, its time period of vibration will be [2015]

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Solution

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 [2015]

1 $\frac{β^{2}}{α}$

2 $\frac{2 π β}{α}$

3 $\frac{β^{2}}{α^{2}}$

4 $\frac{α}{β}$

Ans.
(2)

If $A$ and $ω$ be the amplitude and angular frequency of vibration, then

$α = ω^{2} A$ ...(i)

and $β = ω A$ ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

$\frac{α}{β} = \frac{ω^{2} A}{ω A} = ω$

$∴$ $Time period of vibration is T = \frac{2 π}{ω} = \frac{2 π}{(α / β)} = \frac{2 π β}{α}$

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