At a moment in a progressive wave , the phase of a particle executing S. H. M `(pi)/(3)`. Then the phase of the particle 15 cm ahead and at the time `(T)/(2)` will be , if the wavelength is 60 cm

Answer

To solve the problem, we need to determine the phase of a particle in a progressive wave at a specific position and time. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial phase of the particle, \( \phi_1 = \frac{\pi}{3} \)
- Distance ahead, \( \Delta x = 15 \, \text{cm} \)
- Time increment, \( \Delta t = \frac{T}{2} \)
- Wavelength, \( \lambda = 60 \, \text{cm} \)

2. **Determine the Relationship Between Phase and Position**:
The change in phase \( \Delta \phi \) due to a change in position \( \Delta x \) is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]

3. **Calculate \( \Delta \phi \)**:
Substitute \( \Delta x = 15 \, \text{cm} \) and \( \lambda = 60 \, \text{cm} \) into the equation:
\[ \Delta \phi = \frac{2\pi}{60} \times 15 = \frac{2\pi \times 15}{60} = \frac{2\pi}{4} = \frac{\pi}{2} \]

4. **Determine the New Phase \( \phi_2 \)**:
The new phase \( \phi_2 \) can be calculated using:
\[ \phi_2 = \phi_1 + \Delta \phi \]
Substitute \( \phi_1 = \frac{\pi}{3} \) and \( \Delta \phi = \frac{\pi}{2} \):
\[ \phi_2 = \frac{\pi}{3} + \frac{\pi}{2} \]

5. **Find a Common Denominator**:
To add \( \frac{\pi}{3} \) and \( \frac{\pi}{2} \), we need a common denominator:
- The least common multiple of 3 and 2 is 6.
- Rewrite the fractions:
\[ \frac{\pi}{3} = \frac{2\pi}{6}, \quad \frac{\pi}{2} = \frac{3\pi}{6} \]

6. **Add the Phases**:
Now add the two fractions:
\[ \phi_2 = \frac{2\pi}{6} + \frac{3\pi}{6} = \frac{5\pi}{6} \]

### Final Answer:
The phase of the particle 15 cm ahead and at the time \( \frac{T}{2} \) is:
\[ \phi_2 = \frac{5\pi}{6} \]