A particle starts oscillating simple harmonically from its equilibrium position with time period 'T'. What…

A particle starts oscillating simple harmonically from its equilibrium position with time period 'T'. What is the ratio of potential energy to kinetic energy of the particle at time $t=\frac{T}{12}$ ? $\left(\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}\right)$
  1. $1: 2$
  2. $2: 1$
  3. $1: 3$
  4. $3: 1$

Solution

For SHM, $\begin{aligned} & \text { P.E. }=\frac{1}{2} m \omega^2 x^2 ...(i)\\ & \text { and K.E. }=\frac{1}{2} m \omega^2\left(A^2-x^2\right)...(ii) \end{aligned}$
Given $\mathrm{t}=\frac{\mathrm{T}}{12}$ $\begin{aligned} \therefore \quad x & =A \sin \omega t=A \sin \frac{2 \pi}{T} \cdot t \\ \therefore \quad x & =A \sin \frac{2 \pi}{T} \times \frac{T}{12} \\ & =A \sin \left(\frac{\pi}{6}\right) \\ & =\frac{A}{2}...(iii) \end{aligned}$ Substituting equation (iii) in equation (i), $\begin{aligned} \text { P.E. } & =\frac{1}{2} \mathrm{~m} \omega^2\left(\frac{\mathrm{~A}^2}{4}\right) \\ & =\frac{\mathrm{m} \omega^2 \mathrm{~A}^2}{8} \end{aligned}$
Substituting equation (iii) in equation (ii), $\begin{aligned} \text { K.E. } & =\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\frac{\mathrm{A}^2}{4}\right) \\ & =\frac{1}{2} \mathrm{~m} \omega^2\left(\frac{3 \mathrm{~A}^2}{4}\right) \\ \therefore \quad \frac{\text { P.E. }}{\text { K.E. }} & =\frac{\frac{1}{2} m \omega^2 A^2 \times \frac{1}{4}}{\frac{1}{2} m \omega^2 A^2 \times \frac{3}{4}}=\frac{1}{3} \end{aligned}$ .

Asked in: MHT CET 2024 (10 May Shift 2)

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