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 \([\sin 30^{\circ }=\cos 60^{\circ }=0.5,\ \cos 30^{\circ }=\sin 60^{\circ }=\frac{\sqrt{3}}{2}]\)

• A. \(1:2\)
• B. \(1:3\)
• C. \(2:1\)
• D. \(3:1\)

Hint:

Find the displacement at \(t=T/12\) from \(x=A\sin\omega t\), then use \(U\propto x^{2}\) and \(K\propto A^{2}-x^{2}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Starting from the mean position, \(x=A\sin\omega t\) with \(\omega=\dfrac{2\pi}{T}\).

Step 2: Displacement at \(t=T/12\)
\(\omega t=\dfrac{2\pi}{T}\times\dfrac{T}{12}=\dfrac{\pi}{6}=30^{\circ}\), so \(x=A\sin30^{\circ}=\dfrac{A}{2}\).

Step 3: Calculation
\(U\propto x^{2}=\dfrac{A^{2}}{4}\) and \(K\propto A^{2}-x^{2}=\dfrac{3A^{2}}{4}\).
\(\dfrac{U}{K}=\dfrac{1}{3}\)

Final Answer:
The ratio of potential to kinetic energy is 1:3. \[ \boxed{\text{(B) }1:3} \]