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(B) The potential energy of a simple harmonic oscillator is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$.Potential energy is

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$2$

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(B) The potential energy of a simple harmonic oscillator is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$.Potential energy is maximum when $\sin^2(\omega t) = 1$, which occurs at the extreme positions.For a particle starting from the mean position ($x=0$ at $t=0$), it reaches the first extreme position $(x=A)$ at time $t = T/4$.Given that the maximum potential energy occurs at $t = T / (2 \beta)$, we equate the two expressions:$T / (2 \beta) = T / 4$.By comparing the denominators, we get $2 \beta = 4$, which implies $\beta = 2$.

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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