Question 275: Oscillations

Question

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 $t=\frac{T}{12}$ ? $$\left(\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}\right)$$

Accepted Answer

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

Answer

$1: 2$

Answer

$2: 1$

Answer

$1: 3$

Answer

$3: 1$

Oscillations

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)

Choose the correct answer: