Q 11-13-063JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
A particle of mass $0.50\ \text{kg}$ executes simple harmonic motion under the force $F = -50\ (\text{N m}^{-1})\,x$. The time period of oscillation is $\dfrac{x}{35}\ \text{s}$. The value of $x$ is ______. (Given $\pi = \dfrac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 22
$$T = 2\pi\sqrt{\frac mk} = 2\pi\sqrt{\frac{0.5}{50}} = 0.2\pi = 0.2\times\frac{22}{7} = \frac{22}{35}\ \text{s}$$
So $x = 22$.
Solution by Sreeraj P, M.Sc Physics