Q 11-13-045JEE MainJEE Main 2026 (28 Jan, Shift 1)Easy
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 ______ .
Numerical value type. Enter your answer.
Answer: 2
Potential energy $U = \frac12kx^2$ is maximum at the extreme position, $x = A$:
$$\sin\omega t = 1\Rightarrow \omega t = \frac\pi2\Rightarrow t = \frac{\pi}{2}\cdot\frac{T}{2\pi} = \frac T4$$
(the first such instant). $\dfrac{T}{2\beta} = \dfrac T4\Rightarrow\beta = 2$.
Solution by Sreeraj P, M.Sc Physics