The graph between instantaneous velocity and angular displacement of a particle performing S.H.M. is

Options

A. parabola
B. straight line
C. sinusoidal
D. circle

Answer

To solve the question regarding the graph between instantaneous velocity and angular displacement of a particle performing Simple Harmonic Motion (S.H.M.), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A particle in S.H.M. can be described by its position as a function of time. The standard equation for position is given by:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Finding Instantaneous Velocity**:
- The instantaneous velocity \( v(t) \) can be found by differentiating the position function with respect to time:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
- This can also be expressed as:
\[ v(t) = v_{\text{max}} \cos(\omega t + \phi) \]
where \( v_{\text{max}} = A \omega \).

3. **Relating Angular Displacement**:
- The angular displacement \( \theta \) can be represented as:
\[ \theta = \omega t + \phi \]
- Thus, we can rewrite the velocity in terms of angular displacement:
\[ v = v_{\text{max}} \cos(\theta) \]

4. **Graphing the Relationship**:
- The relationship \( v = v_{\text{max}} \cos(\theta) \) indicates that the instantaneous velocity is proportional to the cosine of the angular displacement.
- The graph of \( v \) versus \( \theta \) will be a cosine wave.

5. **Characteristics of the Graph**:
- The graph will oscillate between \( v_{\text{max}} \) and \(-v_{\text{max}}\).
- The period of the graph will be \( 2\pi \) (since the cosine function has a period of \( 2\pi \)).
- The maximum value occurs at \( \theta = 0, 2\pi, 4\pi, \ldots \) and the minimum value occurs at \( \theta = \pi, 3\pi, 5\pi, \ldots \).

### Conclusion:
The graph between instantaneous velocity and angular displacement of a particle performing S.H.M. is a cosine wave.