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The displacement-time graph of a particle executing S.H.M. is given in the figure (the sketch is schematic and not to scale). Which of the following statements is or are true for this motion? (A) The force is zero at t=(3T)/4. (B) The acceleration is maximum at t=T. (C) The speed is maximum at t=T/4. (D) The P.E. is equal to the K.E. of the oscillation at t=T/2. [Figure: displacement plotted against time as a cosine starting at its maximum at t=0, crossing zero at t=T/4, reaching its minimum at t=T/2, crossing zero again at t=(3T)/4 and returning to maximum at t=T; the time axis is marked T/4, T/2, (3T)/4, T, (5T)/4.]

Asked in JEE Main 2nd Sept 2nd Shift 2020 · v-x and a-x relations

Figure: v-x and a-x relations
Answer: (3) (A), (B) and (C)

Step-by-step solution

The graph is a cosine, x=A cos ω t, with zeros at T/4 and (3T)/4 and extremes at 0, T/2 and T.

At (3T)/4, x=0 so F=-kx=0 (A true); at T, |x|=A so |a|=ω²A is maximum (B true); at T/4, x=0 so the speed is maximum (C true).

At T/2 the particle is at x=-A, where the energy is all potential, so (D) is false.

Why the other options are wrong

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