The acceleration of a particle in $S.H.M.$ is

$A$ particle of mass $10 \text{ g}$ is executing simple harmonic motion with an amplitude of $0.5 \text{ m}$ and a periodic time of $(\pi / 5) \text{ s}$. The maximum value of the force acting on the particle is ... $N$.

$A$ 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:

If the displacement of a body is given by $x = 3 \cos \left( 2 \pi t + \frac{\pi}{4} \right) \text{ m}$,then the acceleration of the body at $t = 2 \text{ s}$ is

$A$ $0.10\, kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x = (10\,cm)\cos [(10\,rad/s)\,t + \pi /2\,rad]$. What is the maximum acceleration experienced by the block?

Give the direction of velocity and acceleration of an $SHM$ particle.