Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the phase angle $\phi$ determines the state (position and direction of motion) of the particle at any given time.
The phase changes uniformly with time.
Step 2: Key Formula or Approach:
The equation for displacement in SHM can be written as $x(t) = A \sin(\omega t + \phi_0)$.
The total phase at any time $t$ is $\phi(t) = \omega t + \phi_0$.
The phase difference $\Delta \phi$ between two times $t_1$ and $t_2$ is the difference in their phases:
$\Delta \phi = \phi(t_2) - \phi(t_1) = (\omega t_2 + \phi_0) - (\omega t_1 + \phi_0) = \omega (t_2 - t_1)$
The angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$.
Step 3: Detailed Explanation:
Given values:
Time period, $T = 16\text{ s}$
Time interval, $\Delta t = t_2 - t_1 = 4\text{ s} - 2\text{ s} = 2\text{ s}$
First, find the angular frequency $\omega$:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8}\text{ rad/s} \]
Now, calculate the phase difference $\Delta \phi$:
\[ \Delta \phi = \omega \times \Delta t \]
\[ \Delta \phi = \frac{\pi}{8} \times 2 \]
\[ \Delta \phi = \frac{2\pi}{8} = \frac{\pi}{4} \]
Step 4: Final Answer:
The phase difference will be $\frac{\pi}{4}$.