Question

Which of the following curves possibly represent one-dimensional motion of a particle?

• A. A, B and D Only
• B. A, D, and C Only
• C. C, D Only
• D. B, and D Only

Hint:

Analyze each graph to determine if it can represent one-dimensional motion. Consider the physical quantities involved and their relationships.

Answer 1

The Correct Option is A

The problem asks to identify which of the four given curves can possibly represent the one-dimensional motion of a particle. We need to analyze the physical validity of each graph.

Concept Used:

To determine if a graph represents a possible physical motion, we must check for consistency with the fundamental principles of kinematics.

1. Single-Valued Functions: For any motion, a particle can only have one position, one velocity, or one value for any other physical state at a single instant in time. Therefore, any graph plotting a physical quantity against time must be a single-valued function of time.

2. Properties of Distance: The total distance covered by a particle is a scalar quantity that is always non-negative and can never decrease with time. The slope of a distance-time graph represents the speed of the particle, which must also be non-negative.

3. Simple Harmonic Motion (SHM): SHM is a common example of one-dimensional motion. Its characteristics are often represented in graphs.

• Displacement: \( x = A \sin(\omega t + \phi_0) \)
• Velocity: \( v = A\omega \cos(\omega t + \phi_0) \)
• Phase: The term \( (\omega t + \phi_0) \) is the phase, which increases linearly with time.
• The relationship between velocity and displacement is given by \( v = \pm \omega \sqrt{A^2 - x^2} \), which can be rearranged to \( \frac{x^2}{A^2} + \frac{v^2}{(A\omega)^2} = 1 \). This is the equation of an ellipse in the velocity-displacement plane (a phase space diagram).

Step-by-Step Solution:

Step 1: Analysis of Curve (A) - Phase vs. Time

This graph shows the phase of a particle as a linear function of time, passing through the origin. This can be represented by the equation \( \text{Phase} = kt \), where k is a constant. This is a valid representation for an oscillatory motion like SHM, where the phase is given by \( \phi(t) = \omega t + \phi_0 \). If the initial phase \( \phi_0 = 0 \), then the phase increases linearly with time. Since SHM is a one-dimensional motion, this curve is physically possible. Therefore, (A) is a possible representation.

Step 2: Analysis of Curve (B) - Velocity vs. Displacement

This graph shows a circular relationship between velocity and displacement. This represents a phase space diagram for an oscillator. For SHM, the relationship is \( \frac{x^2}{A^2} + \frac{v^2}{(A\omega)^2} = 1 \), which is an ellipse. If the angular frequency \( \omega = 1 \) rad/s and the axes are scaled appropriately, this ellipse becomes a circle. At any given displacement \(x\), the particle can have a positive velocity (moving away from the origin) or a negative velocity (moving towards the origin), which is consistent with the graph. Since SHM is a one-dimensional motion, this curve is physically possible. Therefore, (B) is a possible representation.

Step 3: Analysis of Curve (C) - Velocity vs. Time

This graph shows a circular relationship between velocity and time. If we draw a vertical line on this graph at a specific time t (other than the extremes), it intersects the circle at two points. This implies that the particle has two different values of velocity (one positive and one negative) at the same instant in time. This is physically impossible. A particle's velocity must be uniquely defined at any given time. Therefore, (C) is not a possible representation.

Step 4: Analysis of Curve (D) - Total Distance vs. Time

This graph plots the total distance covered against time. The key properties of total distance are that it can never decrease, and its slope (which represents speed) can never be negative. The given graph shows the distance increasing, then remaining constant (when the particle stops), and then increasing again. At no point does the distance decrease. The curve is also a single-valued function of time. This represents a valid physical situation, for instance, a particle moving with constant speed, then stopping for a while, and then moving again. Therefore, (D) is a possible representation.

Final Computation & Result:

Based on the analysis, the curves that possibly represent one-dimensional motion of a particle are (A), (B), and (D). Curve (C) is physically impossible.

We now look at the given options:

(1) A, B and D only

(2) A, B and C only

(3) A and B only

(4) A, C and D only

The correct combination of possible curves is given in option (1).

The correct answer is (1) A, B and D only.

Answer 2

For option (A) \( \phi = kt + C \) it can be 1D motion eg -> \( x = A \sin \phi \) 

(SHM) For option (B) \( v^2 + x^2 = \) constant yes 1D 

For option (C) time can't be negative Not possible 

For option (D) Possible A, B and D only