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Simple Harmonic Motion question

2008 · Shift 2 · Q64
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Simple Harmonic Motion question

2008 · Shift 2 · Q64

JEE AdvancedPhysicsSimple Harmonic MotionMCQ+4 / −1

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 ×\times× 4 matrix given in the ORS.

Column I Column II
(A) Potential energy of a simple pendulum (y-axis) as a function of displacement (x) axis (P) IIT-JEE 2008 Paper 2 Offline Physics - Simple Harmonic Motion Question 5 English 1
(B) 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 (Q) IIT-JEE 2008 Paper 2 Offline Physics - Simple Harmonic Motion Question 5 English 2
(C) Range of a projectile (y-axis) as a function of its velocity (x-axis) when projected at a fixed angle (R) IIT-JEE 2008 Paper 2 Offline Physics - Simple Harmonic Motion Question 5 English 3
(D) The square of the time period (y-axis) of a simple pendulum as a function of its length (x-axis) (S) IIT-JEE 2008 Paper 2 Offline Physics - Simple Harmonic Motion Question 5 English 4

  1. A
    A →\to→(P, S); B →\to→(Q, S); C →\to→(S); (D) →\to→(Q)
  2. B
    A →\to→(S); B →\to→(Q, S); C →\to→(S); (D) →\to→(Q, S)
  3. C
    A →\to→(P, S); B →\to→(Q); C →\to→(S); (D) →\to→(Q, S)
  4. D
    A →\to→(S); B →\to→(Q, S); C →\to→(S, P); (D) →\to→(Q)
View written solutionFree

Correct answer: A

Step 1: Identify the shapes of graphs in Column II

From the figure/options, the intended graph types are the standard ones:

  • PPP: upward-opening curved graph through origin/parabola-type
  • QQQ: straight line through origin
  • RRR: oscillatory/sinusoidal-type
  • SSS: increasing curve which can also represent a parabola-like dependence depending on axes context

We will match each physical quantity using its functional dependence.


Step 2: Match item (A)

(A) Potential energy of a simple pendulum vs displacement

For small angular displacement, a simple pendulum performs SHM, and its potential energy is U=12kx2U=\frac{1}{2}kx^2U=21​kx2 where xxx is displacement from mean position.

So UUU vs xxx is a parabola opening upward.

Hence, (A) matches the parabola-like graphs: PPP and SSS.


Step 3: Match item (B)

(B) Displacement vs time for one-dimensional motion at zero or constant acceleration, body moving along positive xxx-direction

For zero acceleration: x=x0+vtx=x_0+vtx=x0​+vt which is a straight line in ttt.

For constant acceleration: x=x0+ut+12at2x=x_0+ut+\frac{1}{2}at^2x=x0​+ut+21​at2 which is a parabola in ttt.

Since the body moves along positive xxx-direction, displacement increases with time; thus both linear and parabolic increasing graphs are possible.

Hence, (B) matches QQQ and SSS.


Step 4: Match item (C)

(C) Range of projectile vs velocity at fixed angle

Range is R=u2sin⁡2θgR=\frac{u^2\sin 2\theta}{g}R=gu2sin2θ​ At fixed angle θ\thetaθ, R∝u2R \propto u^2R∝u2 So range vs velocity is a parabola through origin.

Hence, (C) matches SSS.


Step 5: Match item (D)

(D) Square of time period of a simple pendulum vs length

For a simple pendulum, T=2πlgT=2\pi\sqrt{\frac{l}{g}}T=2πgl​​ Squaring, T2=4π2glT^2=\frac{4\pi^2}{g}lT2=g4π2​l Thus T2∝lT^2 \propto lT2∝l.

So the graph is a straight line through origin.

Hence, (D) matches QQQ.


Step 6: Final matching

Thus,

  • (A)→(P,S)(A) \to (P,S)(A)→(P,S)
  • (B)→(Q,S)(B) \to (Q,S)(B)→(Q,S)
  • (C)→(S)(C) \to (S)(C)→(S)
  • (D)→(Q)(D) \to (Q)(D)→(Q)

This corresponds to Option A.


Step 7: Compare with stored correct answer

Stored correct answer: A

Our derived answer: A

So the derived answer agrees with the stored answer.

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