Physics · 2026
JEE Main · 28 January 2026, Shift 1 · Q50
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…
The displacement of a particle, executing simple harmonic motion with time period $\displaystyle T$, is expressed as $\displaystyle x(t)=A \sin \omega t$, where $\displaystyle A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $\displaystyle t=T / 2 \beta$. The value of $\displaystyle \beta$ is $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.