Question

The displacement vs time of a particle executing SHM is shown in figure. The initial phase $\phi$ is

• A. $-\pi < \phi < -\frac{\pi}{2}$
• B. $\pi < \phi < \frac{3\pi}{2}$
• C. $-\frac{3\pi}{2} < \phi < -\pi$
• D. $\frac{2\pi}{2} < \phi < \pi$

Answer 1

Answer: (B)

$\pi < \phi < \frac{3\pi}{2}$

Answer 2

The general equation for the displacement of a particle executing SHM can be written as:

$x(t) = A \cos(\omega t + \phi)$

From the given graph:

1. At $t=0$, the displacement $x(0)$ is negative. Substituting $t=0$ into the equation: $x(0) = A \cos(\phi)$ Since $x(0) < 0$ and amplitude $A > 0$, we must have $\cos(\phi) < 0$. This implies that the initial phase $\phi$ must lie in the second or third quadrant (i.e., $\frac{\pi}{2} < \phi < \frac{3\pi}{2}$).

2. At $t=0$, the slope of the $x$ vs $t$ graph is positive, which means the initial velocity $v(0)$ is positive. The velocity $v(t)$ is given by the derivative of $x(t)$ with respect to $t$: $v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$ Substituting $t=0$: $v(0) = -A\omega \sin(\phi)$ Since $v(0) > 0$ and $A\omega > 0$, we must have $-\sin(\phi) > 0$, which means $\sin(\phi) < 0$. This implies that the initial phase $\phi$ must lie in the third or fourth quadrant (i.e., $\pi < \phi < 2\pi$ or $-\pi < \phi < 0$).

Combining both conditions:

• $\cos(\phi) < 0$ (Q2 or Q3)
• $\sin(\phi) < 0$ (Q3 or Q4)

For both conditions to be satisfied simultaneously, the initial phase $\phi$ must be in the third quadrant. Therefore, the range for $\phi$ is $\pi < \phi < \frac{3\pi}{2}$.