JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

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 $\frac{x}{7}$ s, the value of $x$ is __________.

Your answer

Solution

For SHM, $v^{2} = ω^{2} (A^{2} - x^{2})$ . Comparing with the given $v^{2} = 50 - x^{2}$ : $ω^{2} = 1 \Rightarrow ω = 1$ rad/s and $A^{2} = 50$ .

$T = \frac{2 π}{ω} = 2 π$ s.

Set $T = \frac{x}{7}$ : $\frac{x}{7} = 2 π \Rightarrow x = 14 π .$

Numerically: $14 \times 3.14159 \approx 43.98 \approx 44.$

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