Skip to content
Questions 3-23 · Q11

Q.Potential energy of a particle performing linear S.H.M is 0.1π2x20.1\pi^2x^2 joule (x in metre). If mass of the particle is 20 g, find the frequency of S.H.M.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
56% · 46/82 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 →

The potential energy of an S.H.M. particle is Ep=12kx2=12mω2x2E_p=\frac12kx^2=\frac12m\omega^2x^2 (Eq. 5.23). Given Ep=0.1π2x2E_p=0.1\pi^2x^2 (x in metre, energy in joule), comparing coefficients: 12mω2=0.1π2\frac12m\omega^2=0.1\pi^2, so ω2=0.2π2m\omega^2=\frac{0.2\pi^2}{m}. With m=20m=20 g =0.02=0.02 kg, ω2=0.2π20.02=10π2\omega^2=\frac{0.2\pi^2}{0.02}=10\pi^2, so $\omega=\pi\sqrt{10}\approx9. …

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.