Shm Displacement Equation

Reading the general SHM displacement form $x=A\sin (\omega t+\varphi )$ described in NCERT Class 11, Chapter 14 (Oscillations), the coefficient of $t$ inside the sine gives the angular frequency $\omega$. Comparing with $x=4\sin (2\pi t+\pi /6)$, we identify $\omega =2\pi$ rad/s and phase constant $\varphi =\pi /6$. The time period relates to angular frequency by $T=\frac{2\pi }{\omega }$. Substituting, $T=\frac{2\pi }{2\pi }=1$ s. The amplitude 4 m and phase $\pi /6$ do not affect the period, since the period depends only on how fast the phase advances. The option 2 s wrongly uses $\omega =\pi$ by ignoring the factor of 2. The option 0.5 s wrongly doubles the frequency. The option 4 s mistakenly reads the amplitude as the period. It is worth stressing why amplitude and phase are irrelevant here: amplitude $A$ only sets how far the body swings, while the phase constant $\varphi$ only fixes the starting position at $t=0$; neither alters how rapidly the phase angle inside the sine advances, and it is that rate alone that determines the period. One can also cross-check through the frequency: $f=\omega /2\pi =2\pi /2\pi =1$ Hz, and since $T=1/f$, the period is again $1$ s. A dimensional and magnitude check: $\omega$ has units rad/s, and $2\pi /\omega$ gives seconds; a full oscillation in 1 s is physically ordinary for such motion, confirming the answer.