Shm Displacement Equation

Question

The displacement of an oscillating body is written as x equals 4 sin of two pi t plus pi by six in metres, so what is its time period?

RankGuru · Accepted Answer

Reading the general SHM displacement form $x = A\sin(\omega t + \phi)$ 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 $\phi = \pi/6$. The time period relates to angular frequency by $T = \dfrac{2\pi}{\omega}$. Substituting, $T = \dfrac{2\pi}{2\pi} = 1$ 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 $\phi$ 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.

Shm Displacement Equation

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