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 $10 \text{ g}$ 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:

Under the action of a force $F = -75 y$, where $F$ is in Newton and $y$ is in meters, an object of mass $3 \,kg$ executes simple harmonic motion. If the velocity of the object at the mean position is $2.5 \,ms^{-1}$, the maximum acceleration of the object is (in $\,ms^{-2}$)

$A$ particle is performing $S.H.M.$ starting from the extreme position. The graphical representation shows that between displacement and acceleration,there is a phase difference of:

The variation of the acceleration $a$ of a particle executing $S.H.M.$ with displacement $x$ is as shown in the figure. Which of the following graphs correctly represents this relationship?