PROGRESSIVE WAVES

Options

A. phase difference between displacement and accleration of particle is zero
B. Phase difference bwteen displacement and acceleration of particle is `pi`
C. phase difference between between displacement and velocity of particle is `pi//2`
D. phase difference between velocity and acceleration of particle is `pi//2`

Answer

To solve the question regarding the phase differences in progressive waves, we will analyze the relationships between displacement, velocity, and acceleration in simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understanding the Wave Equation**:
The displacement of a particle in SHM can be represented as:
\[ x(t) = A \sin(\omega t) \]
where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(t\) is time.

**Hint**: Identify the basic form of SHM and the variables involved.

2. **Finding the Velocity**:
The velocity \(v(t)\) is the time derivative of displacement \(x(t)\):
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) \]

**Hint**: Remember that velocity is the derivative of displacement.

3. **Finding the Acceleration**:
The acceleration \(a(t)\) is the time derivative of velocity \(v(t)\):
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]

**Hint**: Acceleration is the derivative of velocity.

4. **Determining Phase Differences**:
- **Phase Difference between Displacement and Velocity**:
- Displacement \(x(t) = A \sin(\omega t)\) and Velocity \(v(t) = A \omega \cos(\omega t)\).
- The cosine function leads the sine function by \(\frac{\pi}{2}\) radians.
- Thus, the phase difference between displacement and velocity is:
\[ \Delta \phi_{xv} = \frac{\pi}{2} \text{ radians} \]

**Hint**: Recall the relationship between sine and cosine functions.

- **Phase Difference between Velocity and Acceleration**:
- Velocity \(v(t) = A \omega \cos(\omega t)\) and Acceleration \(a(t) = -A \omega^2 \sin(\omega t)\).
- Again, the cosine function leads the sine function by \(\frac{\pi}{2}\) radians.
- Thus, the phase difference between velocity and acceleration is:
\[ \Delta \phi_{va} = \frac{\pi}{2} \text{ radians} \]

**Hint**: Use the same reasoning as before for the relationship between velocity and acceleration.

- **Phase Difference between Displacement and Acceleration**:
- Displacement \(x(t) = A \sin(\omega t)\) and Acceleration \(a(t) = -A \omega^2 \sin(\omega t)\).
- The acceleration is \(-A \omega^2 \sin(\omega t)\), which means it is in the opposite direction to displacement.
- Therefore, the phase difference between displacement and acceleration is:
\[ \Delta \phi_{xa} = \pi \text{ radians} \]

**Hint**: Recognize that a negative sign indicates a phase shift of \(\pi\).

5. **Conclusion**:
- The phase differences are:
- Between displacement and velocity: \(\frac{\pi}{2}\)
- Between velocity and acceleration: \(\frac{\pi}{2}\)
- Between displacement and acceleration: \(\pi\)

Thus, the correct options based on the phase differences are:
- Displacement and acceleration: \(\pi\) (correct)
- Displacement and velocity: \(\frac{\pi}{2}\) (correct)
- Velocity and acceleration: \(\frac{\pi}{2}\) (correct)

### Final Answer:
The correct options are:
- Phase difference between displacement and acceleration: \(\pi\)
- Phase difference between displacement and velocity: \(\frac{\pi}{2}\)
- Phase difference between velocity and acceleration: \(\frac{\pi}{2}\)