`S_(1)`: The particles speed can never be equal to the wave speed in sine wave if the amplitude is less then wavelength divided by `2pi`.
`S_(2)`: In transverse wave of amplitude A, the maximum particle velocity is four times its wave velocity. Then, the wave length of the wave is `piA`
`S_(3)`: the phase difference between two points separated by 1m in a wave of frequency 120 Hz is `90^(@)`. the velocity of the wave is 480 m/s

Options

A. FTT
B. TTF
C. TFT
D. FFF

Explanation

To solve the problem, we will analyze each of the three statements (S1, S2, and S3) one by one and determine their validity.

### Statement S1:
**Statement:** The particle's speed can never be equal to the wave speed in a sine wave if the amplitude is less than the wavelength divided by \(2\pi\).

**Solution:**
1. **Understanding Particle Speed and Wave Speed:**
- The particle speed \(v_p\) for a sine wave is given by:
\[ v_p = A \omega \]
- The wave speed \(v_w\) is given by:
\[ v_w = \frac{\omega}{k} \]
- Where \(k\) (the wave number) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
- Thus, the wave speed can be rewritten as:
\[ v_w = \frac{\omega \lambda}{2\pi} \]

2. **Condition for Equality:**
- For the particle speed to equal the wave speed, we set:
\[ A \omega = \frac{\omega \lambda}{2\pi} \]
- This simplifies to:
\[ A = \frac{\lambda}{2\pi} \]

3. **Analyzing the Statement:**
- The statement claims that if \(A < \frac{\lambda}{2\pi}\), then \(v_p\) can never equal \(v_w\). This is true because if the amplitude is less than \(\frac{\lambda}{2\pi}\), the particle speed cannot reach the wave speed.

**Conclusion:** Statement S1 is **True**.

---

### Statement S2:
**Statement:** In a transverse wave of amplitude \(A\), the maximum particle velocity is four times its wave velocity. Then, the wavelength of the wave is \(\pi A\).

**Solution:**
1. **Maximum Particle Velocity:**
- The maximum particle velocity is given by:
\[ V_{max} = A \omega \]

2. **Wave Velocity:**
- The wave velocity is given by:
\[ v_w = f \lambda = \frac{\omega}{2\pi} \lambda \]

3. **Given Condition:**
- The statement claims:
\[ A \omega = 4 v_w \]
- Substituting \(v_w\):
\[ A \omega = 4 \left(\frac{\omega}{2\pi} \lambda\right) \]
- Cancelling \(\omega\) (assuming \(\omega \neq 0\)):
\[ A = \frac{2\lambda}{\pi} \]

4. **Finding Wavelength:**
- Rearranging gives:
\[ \lambda = \frac{\pi A}{2} \]
- The statement claims \(\lambda = \pi A\), which is incorrect.

**Conclusion:** Statement S2 is **False**.

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### Statement S3:
**Statement:** The phase difference between two points separated by 1m in a wave of frequency 120 Hz is \(90^\circ\). The velocity of the wave is 480 m/s.

**Solution:**
1. **Phase Difference and Path Difference:**
- The phase difference \(\Delta \phi\) is given as \(90^\circ = \frac{\pi}{2}\) radians.
- The path difference \(\Delta x\) is given as 1 m.

2. **Relating Phase Difference to Wavelength:**
- The relation between phase difference and path difference is:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- Substituting the known values:
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \cdot 1 \]
- Solving for \(\lambda\):
\[ \lambda = 4 \text{ m} \]

3. **Calculating Wave Speed:**
- The wave speed \(v_w\) is given by:
\[ v_w = f \lambda \]
- Substituting \(f = 120 \text{ Hz}\) and \(\lambda = 4 \text{ m}\):
\[ v_w = 120 \times 4 = 480 \text{ m/s} \]

**Conclusion:** Statement S3 is **True**.

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### Final Summary:
- S1: True
- S2: False
- S3: True