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

$A$ piston is performing $S.H.M.$ in the vertical direction with a frequency of $0.5 \,Hz$. $A$ block of $10 \,kg$ is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is: (in $\,m$)

Two simple harmonic motions of angular frequency $100 \, rad \, s^{-1}$ and $1000 \, rad \, s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is

$A$ particle of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

$A$ particle executes $SHM$ on a straight line path. The amplitude of oscillation is $2 \, cm$. When the displacement of the particle from the mean position is $1 \, cm$,the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of velocity. The frequency of $SHM$ (in $s^{-1}$) is:

$A$ point mass oscillates along the $x$-axis according to the law $x = x_0 \cos(\omega t + \pi/4)$. If the acceleration of the particle is written as $a = A \cos(\omega t + \delta)$,then: