The graph between velocity and displacement of a particle executing S.H.M. of angular frequency 1 rad/s can be (where A is the amplitude of displacement)

Options

A. Circle with radius 2 A
B. Circle with radius A
C. Straight line
D. Ellipse

Answer

To solve the problem of finding the graph between velocity and displacement of a particle executing simple harmonic motion (S.H.M.) with an angular frequency of 1 rad/s, we can follow these steps:

### Step-by-Step Solution:

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

2. **Differentiate to Find Velocity**:
- The velocity \( v \) is the time derivative of displacement:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) \]
- Given that \( \omega = 1 \) rad/s, we can simplify this to:
\[ v(t) = A \cos(t) \]

3. **Relate Displacement and Velocity**:
- We know that \( x = A \sin(t) \) and \( v = A \cos(t) \). To eliminate time \( t \), we can use the Pythagorean identity:
\[ \sin^2(t) + \cos^2(t) = 1 \]
- Substituting \( \sin(t) = \frac{x}{A} \) and \( \cos(t) = \frac{v}{A} \), we get:
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{v}{A}\right)^2 = 1 \]

4. **Rearranging the Equation**:
- Multiplying through by \( A^2 \) gives:
\[ x^2 + v^2 = A^2 \]
- This equation represents a circle in the \( x-v \) plane with a radius of \( A \) and centered at the origin (0,0).

5. **Conclusion**:
- The graph between velocity and displacement of a particle executing S.H.M. is a circle with radius \( A \).

### Final Answer:
The graph between velocity and displacement of a particle executing S.H.M. of angular frequency 1 rad/s is a circle with radius \( A \).

---