If a particle is executing S.H.M. then the graph between its acceleration and velocity is , in general

Options

A. An ellipse
B. A circle
C. A parabola
D. A hyperbola

Answer

To solve the problem of finding the relationship between acceleration and velocity for a particle executing Simple Harmonic Motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Basic Equations of SHM**:
- The displacement \( x \) of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

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

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

4. **Relate Velocity and Acceleration**:
- From the equations for velocity and acceleration, we have:
- \( v = A \omega \cos(\omega t) \)
- \( a = -A \omega^2 \sin(\omega t) \)

5. **Express \( \sin(\omega t) \) and \( \cos(\omega t) \)**:
- From the velocity equation, we can express \( \cos(\omega t) \):
\[ \cos(\omega t) = \frac{v}{A \omega} \]
- From the acceleration equation, we can express \( \sin(\omega t) \):
\[ \sin(\omega t) = -\frac{a}{A \omega^2} \]

6. **Use the Pythagorean Identity**:
- We know that \( \sin^2(\omega t) + \cos^2(\omega t) = 1 \). Substituting the expressions from above:
\[ \left(-\frac{a}{A \omega^2}\right)^2 + \left(\frac{v}{A \omega}\right)^2 = 1 \]
- This simplifies to:
\[ \frac{a^2}{A^2 \omega^4} + \frac{v^2}{A^2 \omega^2} = 1 \]

7. **Rearranging the Equation**:
- Multiply through by \( A^2 \omega^4 \):
\[ a^2 + v^2 \omega^2 = A^2 \omega^4 \]
- This is the equation of an ellipse in the \( a \)-\( v \) plane.

8. **Conclusion**:
- The graph between acceleration \( a \) and velocity \( v \) for a particle in SHM is an ellipse.

### Final Answer:
The graph between acceleration and velocity for a particle executing SHM is an **ellipse**.

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