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Time at which an energy condition holds in SHM (KE = PE, maximum PE)

This type of question, asked 3 times: 2021 Paper 22, 30 Jan 2023, 28 Jan 2026 S1. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 22

    22
    A particle of mass 1 kg1\text{ }kg is hanging from a spring of force constant 100Nm−1100Nm^{- 1}. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period TT. The time when the kinetic energy and potential energy of the system will become equal, is Tx\frac{T}{x}. The value of xx is

  2. Question 2

    30 Jan 2023

    21
    The general displacement of a simple harmonic oscillator is x=Asinωtx = Asin\omega t. Let TT be its time period. The slope of its potential energy (U) - time ( tt ) curve will be maximum when t=Tβt =\frac{T}{\beta}. The value of β\beta is

  3. Question 3

    28 Jan 2026 S1

    21
    The displacement of a particle, executing simple harmonic motion with time period TT, is expressed as x(t)=Asinωtx(t) = Asin\omega t, where AA is the amplitude. The maximum value of potential energy of this oscillator is found at t=T/2βt = T/2\beta. The value of β\beta is ____\_\_\_\_ .