Q 11-13-042JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The velocity of a particle executing simple harmonic motion along $x$-axis is described as $v^2 = 50 - x^2$, where $x$ represents displacement. If the time period of motion is $\dfrac{x}{7}$ s, the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 44
Concept. In SHM $v^2=\omega^2(A^2-x^2)=\omega^2A^2-\omega^2x^2$. Comparing with $v^2=50-x^2$ gives $\omega^2$ from the coefficient of $x^2$.
Step 1. Compare. $\omega^2=1\Rightarrow\omega=1$ rad/s (and $A^2=50$).
Step 2. Period. $T=\dfrac{2\pi}{\omega}=2\pi=2\times\dfrac{22}{7}=\dfrac{44}{7}$ s.
Answer. $x=44$.
Solution by Sreeraj P, M.Sc Physics
