A particle performing simple harmonic motion,
(i)has parabolic velocity-displacement graph.
(ii)has sinusoidal velocity-time graph.
(iii) has ellipticalvelocity-acceleration graph.Choose the correct statements.

Options

A. (ii) and (iii)
B. (i), (ii) and (iii)
C. (i) and (ii)
D. (i) and(iii)

Explanation

To solve the question, we need to analyze the statements regarding a particle performing simple harmonic motion (SHM):

1. **Statement (i)**: A particle performing SHM has a parabolic velocity-displacement graph.
2. **Statement (ii)**: A particle performing SHM has a sinusoidal velocity-time graph.
3. **Statement (iii)**: A particle performing SHM has an elliptical velocity-acceleration graph.

Let's evaluate each statement step by step.

### Step 1: Analyze the Velocity-Displacement Graph
The displacement of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t) \]
Differentiating this with respect to time gives us the velocity:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) \]

To find the relationship between velocity and displacement, we can use the identities of sine and cosine. From the displacement equation, we have:
\[ \sin(\omega t) = \frac{x}{A} \]
And from the velocity equation:
\[ \cos(\omega t) = \sqrt{1 - \sin^2(\omega t)} = \sqrt{1 - \left(\frac{x}{A}\right)^2} \]

Thus, we can express velocity in terms of displacement:
\[ v = A \omega \sqrt{1 - \left(\frac{x}{A}\right)^2} \]

This relationship is not parabolic; instead, it represents an elliptical relationship when plotted in the velocity-displacement graph. Therefore, **Statement (i) is incorrect**.

### Step 2: Analyze the Velocity-Time Graph
From our earlier differentiation, we found that:
\[ v(t) = A \omega \cos(\omega t) \]

The cosine function is sinusoidal, which means that the velocity-time graph of a particle in SHM is indeed sinusoidal. Therefore, **Statement (ii) is correct**.

### Step 3: Analyze the Velocity-Acceleration Graph
Next, we differentiate the velocity to find acceleration:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]

Now, we can express the relationship between velocity and acceleration. We already have:
\[ v = A \omega \cos(\omega t) \]
And:
\[ a = -A \omega^2 \sin(\omega t) \]

Using the relationships of sine and cosine, we can derive:
\[ \sin(\omega t) = -\frac{a}{A \omega^2} \]
\[ \cos(\omega t) = \frac{v}{A \omega} \]

If we square both equations and add them, we get:
\[ \left(-\frac{a}{A \omega^2}\right)^2 + \left(\frac{v}{A \omega}\right)^2 = 1 \]

This represents an elliptical relationship in the velocity-acceleration graph. Therefore, **Statement (iii) is correct**.

### Conclusion
Based on our analysis:
- Statement (i) is incorrect.
- Statement (ii) is correct.
- Statement (iii) is correct.

Thus, the correct statements are (ii) and (iii).

### Final Answer
The correct statements are: (ii) and (iii).

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