$A$ particle is executing $S.H.M.$ Then the graph of acceleration as a function of displacement is

For a particle in $SHM$,if the amplitude of the displacement is $a$ and the amplitude of velocity is $v$,the amplitude of acceleration is

$A$ particle executes harmonic motion with an angular velocity and maximum acceleration of $3.5\, rad/s$ and $7.5\, m/s^2$ respectively. The amplitude of oscillation is .... $m$

$A$ point mass oscillates along the $X$-axis according to the law $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$. If the acceleration of the particle is written as $a=A \cos (\omega t-\delta)$, then

$A$ particle is in linear simple harmonic motion between two points,$A$ and $B$,$10 \; cm$ apart. Take the direction from $A$ to $B$ as the positive direction and give the signs of velocity,acceleration,and force on the particle when it is:
$(a)$ at the end $A$.
$(b)$ at the end $B$.
$(c)$ at the mid-point of $AB$ going towards $A$.
$(d)$ at $2 \; cm$ away from $B$ going towards $A$.
$(e)$ at $3 \; cm$ away from $A$ going towards $B$.
$(f)$ at $4 \; cm$ away from $B$ going towards $A$.

The displacement of a particle moving in $S.H.M.$ at any instant is given by $y = a \sin \omega t$. The acceleration after time $t = \frac{T}{4}$ is (where $T$ is the time period).