JEE Main 2017 · 2 Apr · Shift 1 · Physics Q12

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A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like :

Starts at zero; humps touching zero at T/2 and T

Graph of KE against t: starts at zero at t = 0, with humps returning to zero at T/2 and T.

One wide hump with zero at T

Graph of KE against t: starts at zero, one long hump returning to zero at T.

Starts at a maximum; zero at T/2, maximum again at T

Graph of KE against t: starts at a maximum at t = 0, falls to zero at T/2 and alternates, with T marked.

Starts at a maximum; zero at T/4, maximum at T/2

Answer and worked solution

Official answer: D

Starting at equilibrium, $K=K_{max}\cos^2\omega t$ : maximum at $t=0$ , zero at $\frac T4$ , maximum again at $\frac T2$ . Its period is $\frac T2$ . That is graph (4).

Topics:Energy in SHM

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 12