A particle oscillates along x-axis according to law x = x0 sin^2 (t / 2) where x0 = 1.

Question

A particle oscillates along x-axis according to law \(x = x_0 \sin^2\ (t / 2) \ \text{where}\ x_0 = 1\). Variation of kinetic energy (k) with position (x) is given by graph

Answer 1

Correct option is : (3)

\(x = \sin^2 \frac{t}{2} = \frac{1}{2} - \frac{\cos t}{2} \Rightarrow \left( x - \frac{1}{2}\right) =-\frac{cos \ t}{2}\)

x = 0, 1 are the extremes and \(x = \frac{1}{2 }\) is the mean position.