The time period of an oscillator is `8` sec. the phase difference from `t = 2` sec to `l = 4` sec will be :-

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `2pi`

Explanation

To find the phase difference of an oscillator from \( t = 2 \) seconds to \( t = 4 \) seconds, given that the time period \( T \) is \( 8 \) seconds, we can follow these steps:

### Step 1: Determine the angular frequency
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given time period \( T = 8 \) seconds:
\[ \omega = \frac{2\pi}{8} = \frac{\pi}{4} \, \text{rad/sec} \]

### Step 2: Calculate the phase at \( t = 2 \) seconds
The phase \( \phi_1 \) at \( t = 2 \) seconds can be calculated using:
\[ \phi_1 = \omega t_1 = \frac{\pi}{4} \cdot 2 = \frac{\pi}{2} \, \text{radians} \]

### Step 3: Calculate the phase at \( t = 4 \) seconds
The phase \( \phi_2 \) at \( t = 4 \) seconds is:
\[ \phi_2 = \omega t_2 = \frac{\pi}{4} \cdot 4 = \pi \, \text{radians} \]

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) between \( t = 2 \) seconds and \( t = 4 \) seconds is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = \pi - \frac{\pi}{2} = \frac{\pi}{2} \, \text{radians} \]

### Conclusion
The phase difference from \( t = 2 \) seconds to \( t = 4 \) seconds is \( \frac{\pi}{2} \) radians.

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