The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A sin ωt, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T / 2β. The value of β is
Q. The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A sin ωt, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T / 2β. The value of β is ____ .

Correct Answer: 2

Explanation

In simple harmonic motion, the potential energy of the particle is given by:

U = ½ kx²

The potential energy is maximum when the displacement x is maximum.

Given:

x(t) = A sin ωt

Maximum displacement occurs when:

sin ωt = ±1

This happens at:

ωt = π / 2

Using ω = 2π / T:

(2π / T) t = π / 2

t = T / 4

Comparing with:

t = T / 2β

T / 2β = T / 4

β = 2

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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