Question

A particle executes SHM of a amplitude A along X-axis. At t = 0, the position of the particle is $x=\frac{A}{2}$ and it moves along positive X-axis. The displacement of particle in time t is $x=A \sin (\omega t + \delta)$, then the value of phase constant $(\delta)$ will be

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{5\pi}{6}$

Answer 1

Answer: (A)

$\frac{\pi}{6}$

Answer 2

The displacement is given by $x = A \sin(\omega t + \delta)$. At $t=0$, $x = A/2$, so $A/2 = A \sin(\delta)$, which gives $\sin(\delta) = 1/2$. This implies $\delta = \pi/6$ or $\delta = 5\pi/6$.

The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \delta)$. At $t=0$, the particle moves along the positive X-axis, so $v(0) > 0$. $v(0) = A \omega \cos(\delta) > 0$. Since $A$ and $\omega$ are positive, $\cos(\delta) > 0$.

For $\delta = \pi/6$, $\cos(\pi/6) = \sqrt{3}/2 > 0$. This is consistent. For $\delta = 5\pi/6$, $\cos(5\pi/6) = -\sqrt{3}/2 < 0$. This is inconsistent.

Therefore, the phase constant is $\delta = \pi/6$.