Which relations are always correct in between the displacement `(vecx)` velocity `(vecv)` and acceleration `(veca)` for a simple harmonic motion ?

Options

A. `veca*vecx lt0`
B. `vecaxx vecx =0`
C. `vecv*vecx=0`
D. `vecvxx vecx=0`

Answer

To find the correct relationships between displacement \(\vec{x}\), velocity \(\vec{v}\), and acceleration \(\vec{a}\) in simple harmonic motion (SHM), we can analyze the mathematical expressions for each of these quantities.

### Step-by-Step Solution:

1. **Define Displacement in SHM**:
The displacement \(\vec{x}\) in simple harmonic motion can be expressed as:
\[ \vec{x} = A \sin(\omega t) \]
where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(t\) is time.

**Hint**: Remember that displacement in SHM varies sinusoidally with time.

2. **Calculate Velocity**:
The velocity \(\vec{v}\) is the time derivative of displacement:
\[ \vec{v} = \frac{d\vec{x}}{dt} = \frac{d}{dt}(A \sin(\omega t)) = A \omega \cos(\omega t) \]
This can also be rewritten using a phase shift:
\[ \vec{v} = A \omega \sin\left(\omega t + \frac{\pi}{2}\right) \]

**Hint**: The velocity is the derivative of displacement, leading to a cosine function.

3. **Calculate Acceleration**:
The acceleration \(\vec{a}\) is the time derivative of velocity:
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(A \omega \cos(\omega t)) = -A \omega^2 \sin(\omega t) \]
This can also be expressed as:
\[ \vec{a} = -\omega^2 \vec{x} \]

**Hint**: Acceleration is the derivative of velocity, resulting in a negative sine function.

4. **Analyze the Relationships**:
- **Dot Product**:
- For \(\vec{a} \cdot \vec{x}\):
\[ \vec{a} \cdot \vec{x} = |\vec{a}| |\vec{x}| \cos(180^\circ) = -|\vec{a}| |\vec{x}| \]
This shows that \(\vec{a} \cdot \vec{x} < 0\).

- **Cross Product**:
- For \(\vec{a} \times \vec{x}\):
\[ \vec{a} \times \vec{x} = |\vec{a}| |\vec{x}| \sin(180^\circ) = 0 \]
This indicates that \(\vec{a} \times \vec{x} = 0\).

- **Dot Product of Velocity and Displacement**:
- For \(\vec{v} \cdot \vec{x}\):
\[ \vec{v} \cdot \vec{x} = |\vec{v}| |\vec{x}| \cos(90^\circ) = 0 \]
Thus, \(\vec{v} \cdot \vec{x} = 0\).

- **Cross Product of Velocity and Displacement**:
- For \(\vec{v} \times \vec{x}\):
\[ \vec{v} \times \vec{x} = |\vec{v}| |\vec{x}| \sin(90^\circ) \neq 0 \]
This shows that \(\vec{v} \times \vec{x} \neq 0\).

5. **Conclusion**:
From the analysis, we conclude:
- \(\vec{a} \cdot \vec{x} < 0\)
- \(\vec{a} \times \vec{x} = 0\)
- \(\vec{v} \cdot \vec{x} = 0\)
- \(\vec{v} \times \vec{x} \neq 0\)

Thus, the correct relations are:
- A: \(\vec{a} \cdot \vec{x} < 0\)
- B: \(\vec{a} \times \vec{x} = 0\)
- C: \(\vec{v} \cdot \vec{x} = 0\)