For a particle executing SHM along a straight line (displacement is measured from mean position).
`{:("Column - I","Column - II"),("(a) velocity-time graph will be","(p) straight line"),("(b) acceleration-velocity graph may be","(q) circle"),("(c) acceleration-displacement graph will be","(r) ellipse"),("(d) acceleration-time graph will be","(s) sinusoidal curve"):}`

Options

A. a - s, b - q, r, c - p, d - s
B. a - p, b - s, r, c - s, d - p
C. a - q, b - s, r, c - p, d - s
D. a - r, b - q, r, c - q, d - s

Answer

To solve the problem of matching the items in Column I with those in Column II based on the characteristics of Simple Harmonic Motion (SHM), we will analyze each part step by step.

### Step 1: Analyze the velocity-time graph
- **Displacement in SHM**: The displacement \( x(t) \) can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
- **Velocity**: The velocity \( v(t) \) is the derivative of displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
- **Graph Shape**: Since \( v(t) \) is a cosine function, the velocity-time graph will be a sinusoidal curve.

**Match**: (a) velocity-time graph will be (s) sinusoidal curve.

### Step 2: Analyze the acceleration-velocity graph
- **Acceleration**: The acceleration \( a(t) \) is the derivative of velocity:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) \]
- **Relationship**: We can express \( a \) in terms of \( v \):
\[ a = -\frac{\omega^2}{A} v \]
Squaring both sides gives:
\[ \frac{a^2}{\omega^4 A^2} + \frac{v^2}{A^2} = 1 \]
- **Graph Shape**: This is the equation of an ellipse.

**Match**: (b) acceleration-velocity graph may be (r) ellipse.

### Step 3: Analyze the acceleration-displacement graph
- **Using SHM relationship**: From \( a = -\omega^2 x \), we can express acceleration in terms of displacement:
\[ a = -\omega^2 x \]
- **Graph Shape**: This is a linear relationship, which represents a straight line.

**Match**: (c) acceleration-displacement graph will be (p) straight line.

### Step 4: Analyze the acceleration-time graph
- **Acceleration as a function of time**: Since \( a(t) = -A \omega^2 \sin(\omega t + \phi) \), it is also a sinusoidal function.
- **Graph Shape**: Therefore, the acceleration-time graph will be a sinusoidal curve.

**Match**: (d) acceleration-time graph will be (s) sinusoidal curve.

### Final Matches
- (a) → (s)
- (b) → (r)
- (c) → (p)
- (d) → (s)

### Summary of Matches
- A → S
- B → R
- C → P
- D → S