Physics Circle home Student Login

Home›Chapters›Oscillations PYQs›Periodic Motion & SHM Basics›Question 10

MHT CET 2026 (13 May, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2026 (13 May, Shift 1)

Q10MHT CET 2026 · 13 MAY · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle oscillates simple harmonically along a straight line with period 88 s and amplitude 828\sqrt{2} m. It starts from mean position then the ratio of the distances travelled by it in the 2nd2^{nd} second and the first second of its motion is (sin⁡45∘=12) (sin⁡90∘=1)\left(\sin 45^\circ = \frac{1}{\sqrt{2}}\right)\ (\sin 90^\circ = 1)

Show answer and solution

Answer: (c)

  1. ω=2π8=π4\omega = \frac{2\pi}{8} = \frac{\pi}{4}, x=Asin⁡πt4x = A\sin\frac{\pi t}{4} (starts at mean).
  2. x(1)=A2x(1) = \frac{A}{\sqrt{2}}, x(2)=Ax(2) = A; motion is one way in both seconds.
  3. 1st second: A2\frac{A}{\sqrt{2}}; 2nd second: A−A2A - \frac{A}{\sqrt{2}}
  4. Ratio =A(1−1/2)A/2=2−1= \frac{A(1 - 1/\sqrt{2})}{A/\sqrt{2}} = \sqrt{2} - 1
  5. So the answer is (c) 2−1\sqrt{2} - 1.
Report a mistake

More questions on Periodic Motion & SHM Basics