A particle undergoes SHM having time period T. What is the time taken by it to move from mean position to half of amplitude ? a) T2 b) T3 c) T6 d) T12

In simple harmonic motion (SHM), the displacement $x$ of a particle as a function of time $t$ is given by: $$x=A\sin (\omega t+\varphi )$$ where $A$ is the amplitude, $\omega =\frac{2\pi }{T}$ is the angular frequency, and $\varphi$ is the phase constant.

Assuming the particle starts from the mean position ($x=0$) at $t=0$, the displacement equation simplifies to: $$x=A\sin (\omega t)$$

We want the time $t$ when the particle reaches half the amplitude, i.e., $x=\frac{A}{2}$: $$\frac{A}{2}=A\sin (\omega t)$$ $$\sin (\omega t)=\frac{1}{2}$$ $$\omega t=\frac{\pi }{6}(\text{since }\sin (\pi /6)=1/2)$$ $$t=\frac{\pi }{6\omega }=\frac{\pi }{6\cdot \frac{2\pi }{T}}=\frac{T}{12}$$

Thus, the time taken to move from the mean position to half the amplitude is $\frac{T}{12}$. The correct answer is (d).