Physics · Simple Harmonic Motion
JEE Main 2026 — 4 April, Morning Shift — Question 21
The velocity of a particle executing SHM along x‑axis is v² = 50 – x², where x is displacement. If the time period of motion is s, the value of x is ______.
Answer: 44
Numerical answer — enter this value.
Step-by-step solution
Compare with v² = ω²(A²-x²) ⇒ ω=1 rad/s, A=√50. T = 2π/ω = 2π ≈ 6.283. Given T = x/7 ⇒ x = 7T = 44 (approx)
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Physics
- Chapter
- Simple Harmonic Motion
- Topic
- Kinematics of SHM, Phase and Energy in SHM