If a graph is plotted between velocity (v) and displacement (y) of a particle executing SHM from mean position, then the nature of the graph is

Options

A. Straight line
B. Parabola
C. Ellipse
D. Hyperbola

Answer

To determine the nature of the graph plotted between velocity (v) and displacement (y) of a particle executing Simple Harmonic Motion (SHM) from the mean position, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the equations of SHM**:
The displacement \( x \) 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. **Find the velocity**:
The velocity \( v \) of the particle is the time derivative of displacement:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]

3. **Relate velocity to displacement**:
We can express \( v \) in terms of \( x \). From the SHM equations, we know:
\[ v^2 = (A \omega)^2 \cos^2(\omega t + \phi) \]
Using the identity \( \cos^2(\theta) = 1 - \sin^2(\theta) \), we can relate \( \cos^2(\omega t + \phi) \) to \( x \):
\[ \cos^2(\omega t + \phi) = 1 - \sin^2(\omega t + \phi) = 1 - \left(\frac{x}{A}\right)^2 \]
Therefore:
\[ v^2 = (A \omega)^2 \left(1 - \left(\frac{x}{A}\right)^2\right) \]
Simplifying gives:
\[ v^2 = A^2 \omega^2 - \omega^2 x^2 \]

4. **Rearranging the equation**:
Rearranging the equation gives:
\[ \frac{v^2}{A^2 \omega^2} + \frac{x^2}{A^2} = 1 \]
This is the standard form of the equation of an ellipse.

5. **Conclusion about the graph**:
Since the equation we derived is in the form of an ellipse, the graph of velocity (v) versus displacement (x) will be an ellipse.

### Final Answer:
The nature of the graph plotted between velocity (v) and displacement (y) of a particle executing SHM is an **ellipse**.

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