Q 11-13-070JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
The velocity of a particle executing SHM varies with displacement ($x$) as $4v^2=50-x^2$. The time period of oscillations is $\dfrac{x}{7}$ s. The value of $x$ is ______. [Take $\pi=\frac{22}{7}$]
Numerical value type. Enter your answer.
Answer: 88
$v^2=\dfrac14(50-x^2)$. Comparing with $v^2=\omega^2(A^2-x^2)$: $\omega^2=\dfrac14$, $\omega=\dfrac12\ \text{rad s}^{-1}$.
$$T=\frac{2\pi}{\omega}=4\pi=4\times\frac{22}{7}=\frac{88}{7}\ \text{s}$$
So $x=88$.
Solution by Sreeraj P, M.Sc Physics