The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=A sin ω t, 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 _ _ _.

Oscillations and Waves: JEE Main 2026 Physics question

JEE Main 2026 · 28 Jan, Morning shift · Q49

JEE Main 202628 Jan, Morning shiftPhysicsMediumOscillations and Waves185× since 2002

The displacement of a particle, executing simple harmonic motion with time period

T

, is expressed as

x(t)=A\sin\omega t

, where

A

is the amplitude. The maximum value of potential energy of this oscillator is found at

t=T/2\beta

. The value of

\beta

is ______.

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2026 Oscillations and Waves Question with Solution (28 Jan Morning shift, Q49) | JEE Arc