A particle oscillates along the x-axis according to the law, x(t)=x₀ sin ²(t/2) where x₀=1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph

Solution

Physics sir · 4 steps

Rewrite sin² with the double-angle formula: it is plain SHM about x = 1/2, not about the origin.

1. Find the mean position. Use sin² θ = (1 - cos 2θ)/2. With x₀ = 1 m this gives x = 1/2 - 1/2 cos t. So the particle does SHM about x = 1/2 with amplitude A = 1/2 and ω = 1 rad s⁻¹, moving between x = 0 and x = 1.
2. Kinetic energy. In SHM, K = 1/2mω²(A² - y²), where y is the displacement from the mean position. Here y = x - 1/2.

   K = 1/2m[1/4 - (x - 1/2)²] = 1/2m x(1 - x)

3. Read the shape. This is a downward-opening parabola in x. It is zero at x = 0 and x = 1, the turning points, and it is largest, K = m/8, at x = 1/2, the mean position.
4. So. The graph is a smooth inverted parabola from 0 to 1 with its peak at x = 1/2, which is option (d).

The idea to keep. When x(t) has sin² or cos², turn it into cos 2θ first. The constant term tells you the mean position. K is greatest there and zero at the two ends.

Answer from the NTA final answer key. Solution by PrepWiser.