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JEE MAIN 2026
28-01-26 S1
Question
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 $\_\_\_\_$。
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Question Tags
JEE Main
Physics
Medium
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