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NCERT Exemplar · Q16

Q.The displacement–time graph of a particle executing S.H.M. is a cosine curve of period TT: the displacement is at its positive maximum at t=0t=0, zero at t=T/4t=T/4, at its negative maximum (most negative) at t=T/2t=T/2 (i.e. 2T/42T/4), zero again at t=3T/4t=3T/4, and back at its positive maximum at t=Tt=T (i.e. 4T/44T/4), continuing on to t=5T/4t=5T/4. Which of the following statement(s) is/are true? (Note: more than one of the given options may be correct.)

(a) The force is zero at t=3T4t=\dfrac{3T}{4}.
(b) The acceleration is maximum at t=4T4t=\dfrac{4T}{4}.
(c) The velocity is maximum at t=T4t=\dfrac{T}{4}.
(d) The P.E. is equal to the K.E. of oscillation at t=T2t=\dfrac{T}{2}.
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Take x(t)=Acos⁡ωtx(t)=A\cos\omega t with ω=2π/T\omega=2\pi/T. In SHM the restoring force and acceleration are proportional to −x-x (largest at the extremes, zero at the mean), while the speed is largest at the mean position (x=0x=0) and zero at the extremes. Checking the stated instants: force is zero at 3T/43T/4, acceleration is maximum at TT, velocity is maximum at T/4T/4 — all true; but at T/2T/2 the particle is at an extreme, so its energy is all potential, making statement (D) false.

Concept and key relations

For x=Acos⁡ωtx=A\cos\omega t:

v=x˙=−Aωsin⁡ωt,a=x¨=−ω2x,F=ma=−mω2x.v=\dot x=-A\omega\sin\omega t,\qquad a=\ddot x=-\omega^2 x,\qquad F=ma=-m\omega^2 x.

So ∣F∣,∣a∣∝∣x∣|F|,|a|\propto|x| (max at extremes, zero at mean) and ∣v∣|v| is max where x=0x=0. Also PE=12mω2x2\text{PE}=\tfrac12 m\omega^2 x^2 and KE=12mω2(A2−x2)\text{KE}=\tfrac12 m\omega^2(A^2-x^2); these are equal only when ∣x∣=A/2|x|=A/\sqrt2.

Evaluate each statement

  • (A) At t=3T4t=\tfrac{3T}{4}: x=Acos⁡(3π/2)=0x=A\cos(3\pi/2)=0, so F=−mω2x=0F=-m\omega^2x=0. True. …

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