Q 11-13-085JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
A simple pendulum with length $100\ \text{cm}$ and bob of mass $250\ \text{g}$ is executing S.H.M. of amplitude $10\ \text{cm}$. The maximum tension in the string is found to be $\dfrac x{40}$ N. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 99
At the lowest point $v_{max}=A\omega=A\sqrt{\dfrac gl}$, so $\dfrac{v_{max}^2}{l}=\dfrac{A^2g}{l^2}=0.01g$.
$$T_{max}=mg+\frac{mv_{max}^2}{l}=0.25\times9.8\times1.01=2.4745\ \text{N}\approx\frac{99}{40}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics