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← JEE Main 2026 · 4 Apr · Shift 1
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JEE Main 2026 · 4 Apr · Shift 1, Physics question 48

The velocity of a particle executing simple harmonic motion along xx-axis is described as v2=50−x2v^2=50-x^2, where xx represents displacement. If the time period of motion is x7\dfrac x7 s, the value of xx is ________.

Answer and worked solution

Official answer: 44

SHM relation. v2=ω2(A2−x2)v^2=\omega^2(A^2-x^2), so ω=1\omega=1 rad/s and T=2π=447T=2\pi=\frac{44}7 s.

Answer: 44