The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.
The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :

Solution

Physics sir · 4 steps

Potential energy goes as x², so it can never go negative and it repeats every half period.

1. Write U. For SHM the potential energy is U = 1/2kx², measured from zero at the mean position. Squaring x wipes out its sign, so U ≥ 0 at every instant.

   U = 1/2kA²sin² ω t = 1/4kA²(1 - cos 2ω t)

2. Read the zeros. At O, A, B and C the displacement is zero, so U = 0 there. Between these points the particle reaches an extreme, x = ± A, and U peaks at 1/2kA².
3. Half the period. The crest and the trough of the x-t graph both give the same positive peak in U. So U has one hump between each pair of zeros, and its period is T/2.
4. So. The correct plot shows positive humps touching zero at O, A, B and C. That is option (d).

The idea to keep. Energy in SHM depends on x² or v², so it is never negative and it repeats at twice the frequency of the motion. Potential energy is zero at the mean position and maximum at the extremes.

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