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: Understand the relationship between time and phase in SHM
The phase \( \phi \) of an oscillator in simple harmonic motion (SHM) is related to time by the formula:
\[ \phi = \frac{2\pi}{T} \cdot t \]
where \( T \) is the time period and \( t \) is the time.

### Step 2: Calculate the phase at \( t = 2 \) seconds
Using the formula, we can calculate the phase at \( t = 2 \) seconds:
\[ \phi_1 = \frac{2\pi}{8} \cdot 2 = \frac{\pi}{4} \text{ radians} \]

### Step 3: Calculate the phase at \( t = 4 \) seconds
Now, we calculate the phase at \( t = 4 \) seconds:
\[ \phi_2 = \frac{2\pi}{8} \cdot 4 = \frac{\pi}{2} \text{ radians} \]

### Step 4: Find the phase difference
The phase difference \( \Delta \phi \) between the two times is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{2} - \frac{\pi}{4} \]
To simplify this, we can convert \( \frac{\pi}{2} \) to a common denominator:
\[ \Delta \phi = \frac{2\pi}{4} - \frac{\pi}{4} = \frac{\pi}{4} \text{ radians} \]

### Step 5: Final answer
Thus, the phase difference from \( t = 2 \) seconds to \( t = 4 \) seconds is:
\[ \Delta \phi = \frac{\pi}{4} \text{ radians} \text{ or } 45^\circ \]

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