JEE Main 2026 Jan 28 Shift 1, Physics Q21: Simple Harmonic Motion And Oscillation

JEE Main2026Jan 28, Shift 1Physics

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 $t = T /2 β$ . The value of $β$ is ______ .

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Solution

In SHM, potential energy is maximum at the extreme position (maximum displacement). Starting from the mean position at $t = 0$ ( $x = 0$ ), the particle reaches the extreme position after a quarter period:

$t = \frac{T}{4}$ .

Setting $\frac{T}{2 β} = \frac{T}{4} \Rightarrow β = 2$ .

Answer: $2$ .

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JEE Main 2026 Jan 28 Shift 1, Physics Q21: Simple Harmonic Motion And Oscillation

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