JEE Challenger
More from Oscillations

Match Column Parameters with Graph Representations in Physics Experiments

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. Indicate your answer by darkening the appropriate bubbles of the 4×44 \times 4 matrix given in the ORS.

Column I

(A) Potential energy of a simple pendulum (yy-axis) as a function of displacement (xx-axis)

(B) Displacement (yy-axis) as a function of time (xx-axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive xx-direction

(C) Range of a projectile (yy-axis) as a function of its velocity (xx-axis) when projected at a fixed angle

(D) The square of the time period (yy-axis) of a simple pendulum as a function of its length (xx-axis)

Column II

(p)

Diagram 0

(q)

Diagram 1

(r)

Diagram 2

(s)

Diagram 3

Options

A

(A)→(P),  (B)→(Q,S),  (C)→(S),  (D)→(Q,R)(\text{A}) \rightarrow (\text{P}),\; (\text{B}) \rightarrow (\text{Q}, \text{S}),\; (\text{C}) \rightarrow (\text{S}),\; (\text{D}) \rightarrow (\text{Q}, \text{R})

B

(A)→(P,Q),  (B)→(R,S),  (C)→(S),  (D)→(Q)(\text{A}) \rightarrow (\text{P}, \text{Q}),\; (\text{B}) \rightarrow (\text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{S}),\; (\text{D}) \rightarrow (\text{Q})

C

(A)→(P),  (B)→(Q,R,S),  (C)→(Q),  (D)→(S)(\text{A}) \rightarrow (\text{P}),\; (\text{B}) \rightarrow (\text{Q}, \text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{Q}),\; (\text{D}) \rightarrow (\text{S})

D

(A)→(P),  (B)→(Q,R,S),  (C)→(S),  (D)→(Q)(\text{A}) \rightarrow (\text{P}),\; (\text{B}) \rightarrow (\text{Q}, \text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{S}),\; (\text{D}) \rightarrow (\text{Q})

Correct

Step-by-Step Solution

To match the parameters in Column I with the corresponding graph representations in Column II, let us analyze each statement mathematically:


(A) Potential energy of a simple pendulum (yy-axis) as a function of displacement (xx-axis)

For small oscillations, the potential energy UU of a simple pendulum at a displacement xx from its equilibrium position is given by:
U(x)=U0+12mω2x2U(x) = U_0 + \frac{1}{2} m \omega^2 x^2
where U0≥0U_0 \ge 0 is the potential energy at the mean position (x=0x = 0), mm is the mass, and ω\omega is the angular frequency.

  • Graphical representation:
    Here, y=U(x)y = U(x) and xx is the displacement. The relation is quadratic of the form y=c1+c2x2y = c_1 + c_2 x^2 (c2>0c_2 > 0), which represents a parabola opening upwards with its minimum (vertex) located at a non-zero positive value on the yy-axis (y=U0>0y = U_0 > 0).
  • This uniquely matches graph (p).

A→p\mathbf{A \rightarrow p}


(B) Displacement (yy-axis) as a function of time (xx-axis) for a one-dimensional motion at zero or constant acceleration when the body is moving along the positive xx-direction

The position or displacement yy as a function of time tt (here xx-axis) is given by the equation of motion:
y(x)=y0+v0x+12ax2y(x) = y_0 + v_0 x + \frac{1}{2} a x^2

  1. Zero acceleration (a=0a = 0):

    • If the initial displacement y0=0y_0 = 0, then y=v0xy = v_0 x, which is a straight line passing through the origin with a positive slope (v0>0v_0 > 0). This corresponds to graph (q).
    • If the initial displacement y0>0y_0 > 0, then y=y0+v0xy = y_0 + v_0 x, which is a straight line with a positive yy-intercept and a positive slope. This corresponds to graph (r).
  2. Constant positive acceleration (a>0a > 0):

    • If initial position y0=0y_0 = 0 and initial velocity v0=0v_0 = 0, then y=12ax2y = \frac{1}{2} a x^2, which represents a parabola passing through the origin and opening upwards in the first quadrant. This corresponds to graph (s).

B→q,r,s\mathbf{B \rightarrow q, r, s}


(C) Range of a projectile (yy-axis) as a function of its velocity (xx-axis) when projected at a fixed angle

The horizontal range RR of a projectile launched with initial velocity uu at a fixed angle θ\theta to the horizontal is:
R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}

  • Graphical representation:
    Let y=Ry = R and x=ux = u. The relationship becomes:
    y=(sin⁡2θg)x2=Cx2(C>0)y = \left(\frac{\sin 2\theta}{g}\right) x^2 = C x^2 \quad (C > 0)
    This is the equation of a parabola opening upwards and passing through the origin (0,0)(0,0).
  • This corresponds to graph (s).

C→s\mathbf{C \rightarrow s}


(D) The square of the time period (yy-axis) of a simple pendulum as a function of its length (xx-axis)

The time period TT of a simple pendulum of length LL is given by:
T=2πLgT = 2\pi \sqrt{\frac{L}{g}}

Squaring both sides gives:
T2=(4π2g)LT^2 = \left(\frac{4\pi^2}{g}\right) L

  • Graphical representation:
    Let y=T2y = T^2 and x=Lx = L. The relationship becomes:
    y=(4π2g)x=mx(m>0)y = \left(\frac{4\pi^2}{g}\right) x = m x \quad (m > 0)
    This is a straight line passing through the origin (0,0)(0,0) with a positive slope.
  • This corresponds to graph (q).

D→q\mathbf{D \rightarrow q}


Final Match:

  • A →\rightarrow p
  • B →\rightarrow q, r, s
  • C →\rightarrow s
  • D →\rightarrow q