JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Elasticity & SHM

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A particle oscillates along the x-axis according to x(t) = x₀ sin²(πt/2), where x₀ = 1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph: [graph options in source]

(A)[graph - correct: K vs x graph]✓ Correct
(B)[graph - refer to source PDF]
(C)[graph - refer to source PDF]
(D)[graph - refer to source PDF]

Answer: (A)

Solution

Since

x = x_{0} sin^{2} (π t /2) = \frac{x _{0}}{2} [1 - cos (π t)]

, the motion is SHM about

x = x_{0} /2

with amplitude

x_{0} /2 = 0.5 m

and

ω = π

. The kinetic energy

K = \frac{1}{2} m ω^{2} [a^{2} - (x - x_{0} /2)^{2}]

is maximum at the mean position

x = 0.5 m

and zero at the turning points

x = 0

and

x = x_{0} = 1 m

. Hence

K

versus

x

is a downward parabola with roots at

x = 0

and

x = 1 m

and peak at

x = 0.5 m

— graph (A).

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Elasticity & SHM

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