A particle is performing linear S.H.M. If the acceleration and corresponding velocity of the particle are `prop` and v respectively and if we plot `v^2` on Y axis and `prop^2` on X axis, then the graph will be

Options

A. a parabola
B. a straight line
C. an ellipse
D. a hyperbola

Answer

To solve the problem, we need to establish the relationship between the velocity \( v \) and acceleration \( \alpha \) of a particle performing simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understanding SHM**:
- The displacement of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

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

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

4. **Relating \( v \) and \( \alpha \)**:
- From the expressions for \( v \) and \( \alpha \), we can express \( v \) and \( \alpha \) in terms of \( \sin(\omega t + \phi) \) and \( \cos(\omega t + \phi) \):
\[ v = A \omega \cos(\omega t + \phi) \]
\[ \alpha = -A \omega^2 \sin(\omega t + \phi) \]

5. **Squaring Both Expressions**:
- To relate \( v^2 \) and \( \alpha^2 \), we square both equations:
\[ v^2 = (A \omega \cos(\omega t + \phi))^2 = A^2 \omega^2 \cos^2(\omega t + \phi) \]
\[ \alpha^2 = (-A \omega^2 \sin(\omega t + \phi))^2 = A^2 \omega^4 \sin^2(\omega t + \phi) \]

6. **Using the Pythagorean Identity**:
- We know that \( \cos^2(\theta) + \sin^2(\theta) = 1 \). Thus, we can express the relationship:
\[ \frac{v^2}{A^2 \omega^2} + \frac{\alpha^2}{A^2 \omega^4} = 1 \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{v^2}{A^2 \omega^2} + \frac{\alpha^2}{A^2 \omega^4} = 1 \]
- Let \( x = \alpha^2 \) and \( y = v^2 \):
\[ \frac{y}{A^2 \omega^2} + \frac{x}{A^2 \omega^4} = 1 \]

8. **Interpreting the Equation**:
- This is a linear equation in the form of \( \frac{y}{b} + \frac{x}{a} = 1 \), which represents a straight line in the \( v^2 \) vs \( \alpha^2 \) graph.

### Conclusion:
- The graph of \( v^2 \) (on the Y-axis) versus \( \alpha^2 \) (on the X-axis) will be a straight line.