If ' \(\mathrm{v}\) ' is velocity and ' \(a\) ' is acceleration of a particle executing linear simple harmonic motion. Which one of the following statements is correct?

The correct option is (B).
Concept: For SHM, \(\mathrm{a}=-\left(\omega^2\right) \mathrm{x}\) is the necessary condition. A general solution for SHM can be written as,
\(x=A \sin (\omega t+\phi)\)
where, \(\mathrm{x}\) is the displacement from mean position, \(A\) is the amplitude, \(\omega\) is the angular frequency and \(\phi\) is the phase angle.
On taking the first derivative, velocity can be written as
\(
v=\frac{d x}{d t}=(A \omega) \cos (\omega t+f)=(A w)\) \(\sin \left(\omega t+\phi+\frac{\pi}{2}\right)
\)
And, taking derivative of velocity, acceleration can be written as:
\(
\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\left(\mathrm{A} \omega^2\right) \sin (\omega \mathrm{t}+\phi)=\left(\mathrm{A} \omega^2\right)\) \(\sin (\omega \mathrm{t}+\phi+\pi)
\)
Therefore, in SHM acceleration is ahead of velocity by a phase \(\frac{\pi}{2}\) Hence, Option (B) is the only correct option.