Physics Simple Harmonic Motion (Oscillations) JEE Main / Advanced Problems ( Previous Years) MCQ (Single Correct)

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.

Column I

Column II

(a)

Potential energy of a simple pendulum (y–axis) as a function of displacement (x–axis)

(p)

(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)

(c)

Range of a projectile (y–axis) as a function of its velocity (x–axis) when projected at a fixed angle.

(r)

(d)

The square of the time period (y–axis) of a simple pendulum as a function of its length (x–axis).

(s)

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Text Solution

Verified by Experts
The correct answer is:
CHECK THE SOLUTION.

→ (p); → (q, s); → (s); → (q)

Sol. From nature of SHM, the graph of potential energy as function of displacement will be parabolic graph as given in option p. Hence → (p)

a = 0 or a = constant. (as per given condition)

V > 0 moving along positive x-axis

y – displacement

y = ut ± at 2 for a = constant

y = vt for a = 0

These two conditions are satisfied by (q) and (s). → (q, s)

(p) Is rejected because at t = 0 the displacement is not zero and velocity has negative values.

R = and R ∝ u 2

for a fixed angle of projection. at u = 0, R = 0

→ (s)

T =

⇒ T 2 =

⇒ y =

→ (q)

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