The period of SHM of a particle is 12 s. The phase difference between the positions at t = 3 s and t = 4 s will be

Options

A. `pi//4`
B. `3 pi//5`
C. `pi//6`
D. `pi//2`

Explanation

To find the phase difference between the positions of a particle in simple harmonic motion (SHM) at \( t = 3 \) s and \( t = 4 \) s, we can follow these steps:

### Step 1: Determine the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Given that the period \( T = 12 \) s, we can calculate \( \omega \):
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \]

### Step 2: Calculate the phase at \( t = 3 \) s
Using the formula for phase \( \phi \) at time \( t \):
\[ \phi_1 = \omega t \]
Substituting \( t = 3 \) s:
\[ \phi_1 = \omega \cdot 3 = \frac{\pi}{6} \cdot 3 = \frac{3\pi}{6} = \frac{\pi}{2} \text{ rad} \]

### Step 3: Calculate the phase at \( t = 4 \) s
Now, we calculate the phase at \( t = 4 \) s:
\[ \phi_2 = \omega \cdot 4 = \frac{\pi}{6} \cdot 4 = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ rad} \]

### Step 4: Find the phase difference
The phase difference \( \Delta \phi \) between the two times is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values we found:
\[ \Delta \phi = \frac{2\pi}{3} - \frac{\pi}{2} \]
To perform this subtraction, we need a common denominator. The least common multiple of 3 and 2 is 6:
\[ \Delta \phi = \left(\frac{2\pi}{3} \cdot \frac{2}{2}\right) - \left(\frac{\pi}{2} \cdot \frac{3}{3}\right) = \frac{4\pi}{6} - \frac{3\pi}{6} = \frac{\pi}{6} \text{ rad} \]

### Final Answer
The phase difference between the positions at \( t = 3 \) s and \( t = 4 \) s is:
\[ \Delta \phi = \frac{\pi}{6} \text{ rad} \]

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