Skip to content
Question of 66

Q.If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 s, find its maximum velocity.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2022Subjective· 2mImportance★★★★★
0% · 0/66 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

v_max = Aomega, where omega = 2pi/T; with A = 0.05 m and T = 2 s, v_max ≈ 0.157 m/s.

For a particle executing simple harmonic motion with amplitude A and angular frequency omega, the displacement is x = A sin(omega t), so the velocity is v = Aomegacos(omega t), which has a maximum magnitude of:

v_max = A * omega

Given: Amplitude A = 50 mm = 0.05 m

Time period T = 2 s

Angular frequency:

omega = 2pi / T = 2pi / 2 = pi rad/s

Maximum velocity: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.