Q74 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
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 $\_\_\_\_$.
Correct answer: 2
Solution
For SHM with $x(t) = A\sin(\omega t)$, the potential energy is
$U = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t)$.
Maximum potential energy occurs when $\sin^2(\omega t) = 1$, i.e., when $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, ...$
The first maximum occurs at $\omega t = \frac{\pi}{2}$, giving $t = \frac{\pi}{2\omega} = \frac{\pi}{2} \cdot \frac{T}{2\pi} = \frac{T}{4}$.
Since this is given as $t = \frac{T}{2\beta}$.
We have $\frac{T}{4} = \frac{T}{2\beta}$, so $\beta = 2$.
$U = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t)$.
Maximum potential energy occurs when $\sin^2(\omega t) = 1$, i.e., when $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, ...$
The first maximum occurs at $\omega t = \frac{\pi}{2}$, giving $t = \frac{\pi}{2\omega} = \frac{\pi}{2} \cdot \frac{T}{2\pi} = \frac{T}{4}$.
Since this is given as $t = \frac{T}{2\beta}$.
We have $\frac{T}{4} = \frac{T}{2\beta}$, so $\beta = 2$.
Hence, the correct answer is 2.