Question

A particle of mass m is executing SHM with amplitude $A$ and angular frequency $\omega$. The displacement of the particle is given by $x = A \sin(\omega t + \varphi)$. If at $t = 0$, the particle is at $x = A/2$ and moving towards the mean position, the phase constant $\varphi$ is:

• A. $\frac{5\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{2\pi}{3}$

Answer 1

Answer: (A)

$\frac{5\pi}{6}$

Answer 2

The displacement of the particle is given by $x = A \sin(\omega t + \varphi)$. At $t=0$, the displacement is $x = A/2$. Substituting these values into the equation: $A/2 = A \sin(\omega(0) + \varphi)$ $A/2 = A \sin(\varphi)$ $\sin(\varphi) = 1/2$.

The possible values for $\varphi$ in the interval $[0, 2\pi)$ for which $\sin(\varphi) = 1/2$ are $\varphi = \frac{\pi}{6}$ and $\varphi = \frac{5\pi}{6}$.

Next, we use the information about the direction of motion at $t=0$. The particle is moving towards the mean position. The mean position is at $x=0$. The velocity of the particle is given by the derivative of the displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(A \sin(\omega t + \varphi)) = A \omega \cos(\omega t + \varphi)$.

At $t=0$, the velocity is $v(0) = A \omega \cos(\omega(0) + \varphi) = A \omega \cos(\varphi)$.

The particle is at $x = A/2$ at $t=0$. Since $A$ is the amplitude, $A > 0$. Thus, the particle is on the positive side of the mean position ($x=0$). Moving towards the mean position from a positive displacement ($x=A/2$) means the velocity must be negative. So, $v(0) < 0$. $A \omega \cos(\varphi) < 0$. Since $A > 0$ and $\omega > 0$, we must have $\cos(\varphi) < 0$.

Now we check which of the possible values of $\varphi$ ($\frac{\pi}{6}$ and $\frac{5\pi}{6}$) satisfies the condition $\cos(\varphi) < 0$. For $\varphi = \frac{\pi}{6}$: $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. This is positive. This corresponds to moving away from the mean position from $x=A/2$. For $\varphi = \frac{5\pi}{6}$: $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$. This is negative. This corresponds to moving towards the mean position from $x=A/2$.

Therefore, the phase constant $\varphi$ is $\frac{5\pi}{6}$.