Physics Circle home Student Login

Home›Chapters›Oscillations PYQs›Spring Systems›Question 12

MHT CET 2026 (14 May, Shift 1) · Spring Systems

Spring Systems question from MHT CET 2026 (14 May, Shift 1)

Q12MHT CET 2026 · 14 MAY · SHIFT 1Medium

Spring Systems

A mass is suspended from a vertical spring which is executing S.H.M. of frequency 1010 Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (g=10g = 10 m/s2^{2})

Show answer and solution

Answer: (c)

  1. Spring unstretched at the top, so amplitude = static extension: A=mgk=gω2A = \frac{mg}{k} = \frac{g}{\omega^2}.
  2. vmax=Aω=gωv_{max} = A\omega = \frac{g}{\omega}, with ω=2πf=20π\omega = 2\pi f = 20\pi rad/s.
  3. vmax=1020π=12πv_{max} = \frac{10}{20\pi} = \frac{1}{2\pi} m/s
  4. So the answer is (c) 12π\frac{1}{2\pi} m/s.
Report a mistake

More questions on Spring Systems