Question 326: Oscillations

Question

A particle starts oscillating simple harmonically from its mean position with time period '$$T$$'. At time $$t=\frac{T}{12}$$, the ratio of the potential energy to kinetic energy of the particle is $$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$$

Accepted Answer

The particle starts from the mean position.
$$\begin{aligned}
\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t} & =\mathrm{A} \sin \left(\frac{2 \pi}{\mathrm{T}}ight) 	imes \mathrm{t} \
& =A \sin \left(\frac{2 \pi}{\mathrm{T}} 	imes \frac{\mathrm{T}}{12}ight) \
& =A \sin \left(\frac{\pi}{6}ight)
\end{aligned}$$
$$	herefore \mathrm{x}=\mathrm{A} \sin 30^{\circ}=\frac{\mathrm{A}}{2} \quad 	herefore \mathrm{x}^2=\frac{\mathrm{A}^2}{4}$$
$$	herefore$$ The particle is at a distance $$\mathrm{A} / 2$$ from the mean position.
At this point its P.E. $$=\frac{1}{2} K x^2=\frac{1}{2} m \omega^2 x^2\quad 	ext{...... (1)}$$
and its K.E. $$=\frac{1}{2} \mathrm{mv}^2$$
$$	herefore \mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{~A}^2-\mathrm{x}^2ight)\quad 	ext{..... (2)}$$
$$\begin{aligned}
& 	herefore \frac{	ext { P.E. }(\mathrm{U})}{	ext { K.E. }(\mathrm{K})}=\frac{\mathrm{x}^2}{\mathrm{~A}^2-\mathrm{x}^2}=\frac{\frac{\mathrm{A}^2}{4}}{\mathrm{~A}^2-\frac{\mathrm{A}^2}{4}}=\frac{\frac{\mathrm{A}^2}{4}}{\frac{3 \mathrm{~A}^2}{4}} \
& 	herefore \frac{\mathrm{U}}{\mathrm{K}}=\frac{1}{3}
\end{aligned}$$

Answer

$$1: 2$$

Answer

$$3: 1$$

Answer

$$2: 1$$

Answer

$$1: 3$$

Oscillations

A particle starts oscillating simple harmonically from its mean position with time period '$T'. At time t=\frac{T}{12}, the ratio of the potential energy to kinetic energy of the particle is \left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)

Choose the correct answer: