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Q.Show that in the case of simple harmonic motion ω=km\omega = \sqrt{\dfrac{k}{m}}

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2026Subjective· 3mImportance★★★★★
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Spring force −kx-kx gives a=−(k/m)xa = -(k/m)x; matching a=−ω2xa = -\omega^{2}x yields ω=k/m\omega = \sqrt{k/m}.

For a block of mass mm attached to a spring of force constant kk, the restoring force when displaced by xx from the mean position is (Hooke's law)

F=−kx.F = -kx.

By Newton's second law, F=maF = ma, so

ma=−kx⇒a=−kmx.ma = -kx \Rightarrow a = -\frac{k}{m}x.

The defining equation of simple harmonic motion is

a=−ω2x,a = -\omega^{2}x, …

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