Question

Position of a \(3\ \text{kg}\) mass moving along the \(x\)-axis is given by \(x=0.3\cos(\omega t)\ \text{m}\). If \(K(t)\) denotes the kinetic energy at time \(t\), then the value of \(\dfrac{K\left(\frac{\pi}{6\omega}\right)}{K\left(\frac{\pi}{3\omega}\right)}\) is

• A. \(\dfrac{1}{3}\)
• B. \(\dfrac{1}{2}\)
• C. \(\dfrac{\sqrt{3}}{2}\)
• D. \(\sqrt{3}\)

Hint:

For SHM given by \[ x=A\cos(\omega t), \] velocity is \[ v=-A\omega\sin(\omega t), \] so kinetic energy varies as \[ K\propto \sin^2(\omega t). \]

Answer 1

The Correct Option is A

Step 1: Write kinetic energy as a function of time.
\( x = 0.3\cos(\omega t) \Rightarrow v = -0.3\omega\sin(\omega t) \). So \( K(t) \propto \sin^2(\omega t) \).

Step 2: Evaluate the ratio at the two times.
At \( t_1 = \frac{\pi}{6\omega} \): \( \sin(30^\circ) = 1/2 \Rightarrow \sin^2 = 1/4 \). At \( t_2 = \frac{\pi}{3\omega} \): \( \sin(60^\circ) = \sqrt{3}/2 \Rightarrow \sin^2 = 3/4 \). \[ \frac{K(t_1)}{K(t_2)} = \frac{1/4}{3/4} \] \[ \boxed{\dfrac{1}{3}} \]