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JEE Main 2026 (4 Apr, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from JEE Main 2026 (4 Apr, Shift 1)

Q5JEE Main 2026 · 4 APR · SHIFT 1Easy

Velocity & Acceleration in SHM

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 s\frac{x}{7}\text{ s}, the value of xx is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 44

  1. For SHM v2=ω2(A2−x2)=ω2A2−ω2x2v^2 = \omega^2(A^2 - x^2) = \omega^2A^2 - \omega^2x^2
  2. Compare with v2=50−x2v^2 = 50 - x^2: ω2=1⇒ω=1\omega^2 = 1 \Rightarrow \omega = 1 rad/s
  3. T=2πω=2π=2×227=447T = \frac{2\pi}{\omega} = 2\pi = 2\times\frac{22}{7} = \frac{44}{7} s
  4. x7=447⇒x=44\frac{x}{7} = \frac{44}{7} \Rightarrow x = 44
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