Question 231: Oscillations

Question

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $\mathrm{t}=\frac{\mathrm{T}}{6}$, the ratio of the potential energy to kinetic energy of the particle is
$$ \left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right] $$

Accepted Answer

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

Answer

$1: 2$

Answer

$1: 3$

Answer

$2: 1$

Answer

$3: 1$

Oscillations

Choose the correct answer: