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The magnitude of maximum acceleration is $\pi$ times that of maximum velocity of a simple harmonic oscillator. The time period of the oscillator in seconds is
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The correct answer is:
$2$
Maximum acceleration $=$ Maximum velocity $\times \pi$ ie,
$\omega^2 A=\pi \omega A$
where $A$ is amplitude and $\omega$ is angular velocity.
$\begin{aligned} & \Rightarrow \quad \omega=\pi \\ & \Rightarrow \quad \frac{2 \pi}{T}=\pi \Rightarrow T=2 \mathrm{~s} \\ & \end{aligned}$
$\omega^2 A=\pi \omega A$
where $A$ is amplitude and $\omega$ is angular velocity.
$\begin{aligned} & \Rightarrow \quad \omega=\pi \\ & \Rightarrow \quad \frac{2 \pi}{T}=\pi \Rightarrow T=2 \mathrm{~s} \\ & \end{aligned}$
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