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Q.Show that in the case of simple harmonic motion, omega = sqrt(k/m). OR A body oscillates with SHM according to the equation (in SI units) x = 5cos[2pit + pi/4]. Calculate the frequency and rewrite the expression for x at t = 1.5 s.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2023Subjective· 2mImportance★★★★★
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Main: ω=k/m\omega=\sqrt{k/m} follows from F=−kx=maF=-kx=ma. OR: f=1f=1 Hz and x(1.5 s)≈−3.54x(1.5\,\text{s})\approx-3.54 m.

Main part. For a particle undergoing SHM under a Hooke's-law restoring force, F=−kxF=-kx. By Newton's second law, F=maF=ma, so ma=−kxma=-kx, giving acceleration a=−kmxa=-\dfrac{k}{m}x. But by the defining kinematic equation of SHM, a=−ω2xa=-\omega^2x. Comparing the two expressions for aa: −ω2x=−kmx ⇒ ω2=km ⇒ ω=km-\omega^2x=-\dfrac{k}{m}x\ \Rightarrow\ \omega^2=\dfrac{k}{m}\ \Rightarrow\ \omega=\sqrt{\dfrac{k}{m}}.

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