JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Elasticity & SHM

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A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

(A) $9.8 m / s^{2}$
(B) $10 m / s^{2}$
(C) $5 m / s^{2}$
(D) $4 m / s^{2}$ ✓ Correct

Answer: (D)

Solution

The potential energy curve is a parabola with its minimum at the equilibrium and a height of

10 J

at a displacement of

2 m

from it. Thus

\frac{1}{2} k A^{2} = 10

with

A = 2 m

gives

k = 5 N/m

. The spring period is

T = 2 π \frac{m}{k} = 2 π \frac{5}{5} = 2 π s

. Equating to the pendulum period

2 π \frac{ℓ}{g}

with

ℓ = 4 m

gives

\frac{4}{g} = 1

, so

g = 4 m/s^{2}

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Elasticity & SHM

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