A particle executes `SHM` of amplitude `A` and time period `T` The distance travelled by the particle in the during its phasde changes from `(pi)/(12)` to `(5 pi)/(12)` sin15=0.26,sin75=0.96

Options

A. `(1)/(sqrt(2)) A`
B. `sqrt((3)/(2))A`
C. `(2)/sqrt(3)A`
D. `sqrt((2)/(3))A`

Explanation

To solve the problem of finding the distance traveled by a particle executing Simple Harmonic Motion (SHM) as its phase changes from \(\frac{\pi}{12}\) to \(\frac{5\pi}{12}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Phase Angles**:
- The initial phase is \(\phi_1 = \frac{\pi}{12}\).
- The final phase is \(\phi_2 = \frac{5\pi}{12}\).

2. **Convert Phase Angles to Degrees**:
- \(\phi_1 = \frac{\pi}{12} = 15^\circ\)
- \(\phi_2 = \frac{5\pi}{12} = 75^\circ\)

3. **Use the SHM Position Formula**:
- The position of the particle in SHM is given by:
\[ x(t) = A \sin(\phi) \]
- Therefore, the positions at the two phases are:
- \(x_1 = A \sin\left(\frac{\pi}{12}\right)\)
- \(x_2 = A \sin\left(\frac{5\pi}{12}\right)\)

4. **Substitute the Values**:
- From the problem, we know:
- \(\sin\left(\frac{\pi}{12}\right) = \sin(15^\circ) = 0.26\)
- \(\sin\left(\frac{5\pi}{12}\right) = \sin(75^\circ) = 0.96\)
- Thus:
- \(x_1 = A \cdot 0.26\)
- \(x_2 = A \cdot 0.96\)

5. **Calculate the Distance Traveled**:
- The distance traveled by the particle as it moves from phase \(\phi_1\) to \(\phi_2\) is given by:
\[ \text{Distance} = x_2 - x_1 = A \cdot 0.96 - A \cdot 0.26 \]
- Simplifying this:
\[ \text{Distance} = A (0.96 - 0.26) = A \cdot 0.70 \]

6. **Final Result**:
- The distance traveled by the particle during its phase change from \(\frac{\pi}{12}\) to \(\frac{5\pi}{12}\) is:
\[ \text{Distance} = 0.70 A \]