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(B) The maximum acceleration of a simple harmonic oscillator is given by $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the

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$2$

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(B) The maximum acceleration of a simple harmonic oscillator is given by $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.The maximum velocity is given by $v_{max} = \omega A$.According to the problem,the magnitude of maximum acceleration is $\pi$ times the maximum velocity:$a_{max} = \pi \cdot v_{max}$Substituting the expressions:$\omega^2 A = \pi \cdot \omega A$Dividing both sides by $\omega A$ (assuming $\omega, A 
eq 0$):$\omega = \pi$We know that the angular frequency $\omega$ is related to the time period $T$ by the formula $\omega = \frac{2\pi}{T}$.Substituting $\omega = \pi$:$\pi = \frac{2\pi}{T}$$T = 2 	ext{ s}$.

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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