The amplitude of a particle executing $S.H.M.$ with a frequency of $60 \, Hz$ is $0.01 \, m$. The maximum value of the acceleration of the particle is

For a particle performing Simple Harmonic Motion $(S.H.M.)$,the displacement-time graph is as shown. For that particle,the force-time graph is correctly represented by which of the following graphs?

$A$ body oscillates with a simple harmonic motion having amplitude $0.05 \, m$. At a certain instant,its displacement is $0.01 \, m$ and acceleration is $1.0 \, m/s^2$. The period of oscillation 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:

$A$ particle moving along the $X$-axis executes simple harmonic motion. The force acting on it is given by: (Where $A$ and $K$ are positive constants)

$A$ horizontal platform with a small object placed on it executes a linear $S.H.M.$ in the vertical direction. The amplitude of oscillation is $40 \text{ cm}$. What should be the least period of these oscillations, so that the object is not detached from the platform (in $\pi \text{ s}$)? [Take $g = 10 \text{ m/s}^2$]