1. Write the SHM relations

For a particle in simple harmonic motion, $a=-\omega^2 x$ where $\omega=\frac{2\pi}{T}$

Also, velocity and displacement satisfy $v^2=\omega^2(A^2-x^2)$ though we may not need this immediately.

2. Check each option

Option A: $\dfrac{aT}{x}$

Using $a=-\omega^2 x$, $\frac{aT}{x}=\frac{(-\omega^2 x)T}{x}=-\omega^2 T$ Now substitute $\omega=\dfrac{2\pi}{T}$: $-\omega^2 T=-\left(\frac{2\pi}{T}\right)^2 T=-\frac{4\pi^2}{T}$ This is a constant, so A does not change with time.

Option B: $aT+2\pi v$

Substitute $a=-\omega^2 x$: $aT+2\pi v=-\omega^2 x\,T+2\pi v$ Since both $x$ and $v$ vary with time, this is not obviously constant. Let us rewrite using $\omega=\frac{2\pi}{T}\implies \omega T=2\pi$ Then $aT+2\pi v=-\omega^2 xT+2\pi v=-\omega(\omega Tx)+2\pi v=-2\pi\omega x+2\pi v$ $=2\pi(v-\omega x)$ Now in SHM, if $x=A\cos(\omega t+\phi),$ then $v=-A\omega\sin(\omega t+\phi)$ So $v-\omega x=-A\omega\sin(\omega t+\phi)-A\omega\cos(\omega t+\phi)$ which clearly changes with time. Hence B changes with time.

Option C: $\dfrac{aT}{v}$

Using $a=-\omega^2 x$, $\frac{aT}{v}=\frac{-\omega^2 x\,T}{v}$ Since both $x$ and $v$ are time-dependent and their ratio is not constant in SHM, this varies with time.

To confirm, using $x=A\cos(\omega t+\phi),\qquad v=-A\omega\sin(\omega t+\phi),$ we get