1.

Derive expressions for the period of SHM in terms of (1) angular frequency (2) force constant (3) acceleration.

Answer»

The general expression for the displacement (x) of a particle performing SHM is x = A sin (ωt + α)

(1) Let T be the period of the SHM and x the displacement after a further time interval T. Then

x1 = A sin [ω(t + T) + α] 

= A sin (ωt + ωT + α) 

= A sin (ωt + α + ωT)

Since T ≠ 0, for x1 to be equal to x, we must have (ωT)min = 2π.

Hence, the period (T) of SHM is T = 2π/ω This is the expression for the period in terms of the constant co, the angular frequency.

(2) If m is the mass of the particle and k is the force constant, ω = \(\sqrt{k/m}\).

∴T = \(\frac{2\pi}\omega\) = \(\frac{2\pi}{\sqrt{k/m}}\) = 2π \(\sqrt{\frac mk}\)

(3) The acceleration of a particle performing SHM has a magnitude a = ω2x

∴ ω = \(\sqrt{a/x}\)

\(\sqrt{acceleration\,per\,unit\,displacement}\)

∴ T = \(\frac{2\pi}\omega\) = \(\frac{2\pi}{\sqrt{acceleration\,per\,unit\,displacement}}\)



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