The magnitude of maximum acceleration is pi t | vedclass

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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}$$\omega^2 A = \pi \cdot \omega A$Dividing both sides by $\omega A$ (assuming $\omega, A 
eq 0$):$\omega = \pi$Since the angular frequency $\omega = \frac{2\pi}{T}$,where $T$ is the time period:$\frac{2\pi}{T} = \pi$$T = \frac{2\pi}{\pi} = 2 \ s$Therefore,the time period of the oscillator is $2 \ s$.

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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