A simple harmonic motion of a particle is represented as, y = 10 cos ωt 10. The acceleration of the particle at time t = $\frac{\pi} {2\omega}$ will be : (symbols here carry their usual meanings)

The given equation for the simple harmonic motion of a particle is $y = 10 \cos \omega t + 10$. This represents the displacement of the particle from a reference point (in this case, an origin shifted by 10 units). The velocity of the particle is the first derivative of displacement with respect to time: $v = \frac{dy}{dt} = \frac{d}{dt}(10 \cos \omega t + 10) = -10 \omega \sin \omega t$. The acceleration of the particle is the first derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt} (-10 \omega \sin \omega t) = -10 \omega^2 \cos \omega t$. We need to find the acceleration at time $t = \frac{\pi}{2\omega}$. Substituting this value of $t$ into the acceleration equation: $a\left(t=\frac{\pi}{2\omega}\right) = -10 \omega^2 \cos\left(\omega \cdot \frac{\pi} {2\omega}\right) = -10 \omega^2 \cos\left(\frac{\pi}{2}\right)$. Since $ \cos\left(\frac{\pi}{2}\right) = 0$, the acceleration at this time is $a = -10 \omega^2 \cdot 0 = 0$.