Column I gives a list of possible set of parameters measured in some experiments. The variations of

Q: Why does Scenario A match with both (p) and (s)?

The potential energy of a simple pendulum is a quadratic function of displacement, which is represented by a parabola. If the origin is chosen at the mean position, the potential energy is zero at $x=0$, matching graph (s). If the mean position is shifted away from the origin, the minimum of the parabola is shifted, matching graph (p).

Q: Why does Scenario B match with three different graphs (q, r, s)?

Scenario B specifies 'zero or constant acceleration'. If acceleration is zero, the motion is uniform, and displacement is linear with time, matching graph (q). If acceleration is constant and non-zero, displacement is quadratic with time, which can match either graph (r) or graph (s) depending on the magnitude of the acceleration.

Q: Why is the range of a projectile in Scenario C matched with (s) and not (r)?

Both (r) and (s) represent quadratic curves. However, the range $R = \frac{\sin 2\theta}{g} v^2$ typically increases very steeply with velocity $v$ because of the quadratic dependence, which is best represented by the steeper curve in graph (s).

Q: Is the potential energy of a pendulum always a perfect parabola?

No, the potential energy $U = mgL(1 - \cos\theta)$ is only approximately parabolic ($U \approx \frac{mg}{2L}x^2$) for small angular displacements where $\cos\theta \approx 1 - \frac{\theta^2}{2}$ holds true.

Q: Why does $T^2$ vs $L$ in Scenario D give a straight line instead of a curve?

While the time period $T$ is proportional to $\sqrt{L}$ (which would be a curve), the square of the time period $T^2$ is directly proportional to $L$. This linear relationship results in a straight line passing through the origin.

Question

Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II.

List-I:

P. Potential energy of a simple pendulum ($Y$-axis) as a function of displacement ($X$-axis).

Q. Displacement ($Y$-axis) as a function of time ($X$-axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive $x$-direction.

R. Range of a projectile ($Y$-axis) as a function of its velocity ($X$-axis) when projected at a fixed angle.

S. The square of the time period ($Y$-axis) of a simple pendulum as a function of its length ($X$-axis).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In this problem, we are tasked with matching physical scenarios involving simple pendulums, one-dimensional motion, and projectile motion with their corresponding graphical representations.

Let's break down each scenario mathematically to identify the relationship between the dependent variable ($Y$-axis) and the independent variable ($X$-axis).

---

### Scenario A: Potential Energy of a Simple Pendulum

For a simple pendulum of mass $m$ and length $L$, when the bob is displaced by an angle $	heta$, its height relative to the lowest point is:

$$h = L(1 - \cos	heta)$$

Thus, the potential energy $U$ is given by:

$$U = mgh = mgL(1 - \cos	heta)$$

For small angular displacements ($x \ll L$), where $x$ is the linear displacement along the arc:

$$\cos	heta \approx 1 - \frac{	heta^2}{2} = 1 - \frac{x^2}{2L^2}$$

Substituting this approximation back into the potential energy equation yields:

$$U(x) \approx \frac{mg}{2L} x^2$$

This is a quadratic relationship of the form $Y = C X^2$, which represents a **parabola opening upwards**.

* If the mean position is chosen at the origin ($x = 0$), the potential energy is zero at $x = 0$, which corresponds to **graph (s)**.
* If the mean position is shifted away from the origin, the minimum of the potential energy curve shifts, which corresponds to **graph (p)**.

Therefore, **(A) matches with (p) and (s)**.

---

### Scenario B: One-Dimensional Motion

We are given a body moving along the positive $x$-direction under two possible conditions: zero acceleration or constant acceleration.

#### Case 1: Zero Acceleration ($a = 0$)

With zero acceleration, the velocity $v$ is constant. The displacement $s$ as a function of time $t$ is:

$$s(t) = v t$$

This is a linear relationship of the form $Y = m X$, which represents a **straight line passing through the origin with a positive slope**. This corresponds to **graph (q)**.

#### Case 2: Constant Acceleration ($a = 	ext{constant}$)

With a constant acceleration $a$, the displacement $s$ as a function of time $t$ is given by the second equation of motion:

$$s(t) = u t + \frac{1}{2} a t^2$$

This is a quadratic relationship of the form $Y = u X + \frac{1}{2} a X^2$. Since the body is moving along the positive $x$-direction, the curve starts at the origin and curves upwards with an increasing slope. This corresponds to **graph (r)** or **graph (s)**.

Therefore, **(B) matches with (q), (r), and (s)**.

---

### Scenario C: Range of a Projectile

The horizontal range $R$ of a projectile launched with an initial velocity $v$ at a fixed angle $	heta$ is given by:

$$R = \frac{v^2 \sin 2	heta}{g}$$

Since the launch angle $	heta$ and the acceleration due to gravity $g$ are constant, we can write:

$$R = C v^2 \quad 	ext{where} \quad C = \frac{\sin 2	heta}{g}$$

This is a quadratic relationship of the form $Y = C X^2$, which represents a **parabola starting from the origin and curving upwards steeply**. This corresponds to **graph (s)**.

Therefore, **(C) matches with (s)**.

---

### Scenario D: Square of the Time Period of a Simple Pendulum

The time period $T$ of a simple pendulum of length $L$ is given by:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

Squaring both sides of the equation gives:

$$T^2 = \left(\frac{4\pi^2}{g}ight) L$$

Letting $Y = T^2$ and $X = L$, we get:

$$Y = m X \quad 	ext{where} \quad m = \frac{4\pi^2}{g}$$

This is a linear relationship of the form $Y = m X$, which represents a **straight line passing through the origin with a positive slope**. This corresponds to **graph (q)**.

Therefore, **(D) matches with (q)**.

Animated Solution for Physics - Oscillations: Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Understanding the Physical Systems

Potential Energy of a Simple Pendulum

1D Motion with Zero Acceleration ( $a = 0$ )

1D Motion with Constant Acceleration ( $a = const$ )

Range of a Projectile vs Velocity

Square of Time Period of a Pendulum

Final Matrix Matching

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

Scenario A

Scenario B

# Case 1

# Case 2

Scenario C

Scenario D

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Animated Solution for Physics - Oscillations: Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Understanding the Physical Systems

Potential Energy of a Simple Pendulum

1D Motion with Zero Acceleration ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=0)

1D Motion with Constant Acceleration ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=const)

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