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NCERT Exemplar · Q14

Q.The displacement–time curve of a particle executing S.H.M. is a smooth sinusoid of period 4 s4\ \text{s}. The displacement is positive and near its maximum at t=0 st=0\ \text{s}, falls to zero at about t=1 st=1\ \text{s}, reaches its most negative value at about t=2 st=2\ \text{s}, returns to zero at about t=3 st=3\ \text{s}, and rises back to its positive maximum at t=4 st=4\ \text{s}; this 4-second cycle then repeats up to t=7 st=7\ \text{s}. Choose the correct statement(s). (Note: more than one of the given options may be correct.)

(a) Phase of the oscillator is same at t=0t=0 s and t=2t=2 s.
(b) Phase of the oscillator is same at t=2t=2 s and t=6t=6 s.
(c) Phase of the oscillator is same at t=1t=1 s and t=7t=7 s.
(d) Phase of the oscillator is same at t=1t=1 s and t=5t=5 s.
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Same phase means same displacement and same direction of motion, which recurs exactly one full period later. Reading the period as T=4 sT=4\ \text{s}, only time separations equal to 4 s4\ \text{s} (or any multiple) give the same phase. That is true for the pair 2 s,6 s2\text{ s},6\text{ s} and the pair 1 s,5 s1\text{ s},5\text{ s}.

Concept: phase repeats every period

The phase of an SHM advances by 2π2\pi each period. Two instants t1,t2t_1,t_2 are in phase iff

t2−t1=nT,n=0,1,2,…t_2-t_1=nT,\quad n=0,1,2,\dots

If the separation is an odd multiple of T/2T/2, the states are in anti-phase (equal and opposite).

Read the period

The curve completes one full oscillation in 4 s4\ \text{s}, so T=4 sT=4\ \text{s}.

Test each option

  • (A) t=0,2t=0,2: Δt=2 s=T/2\Delta t=2\ \text{s}=T/2 ⇒\Rightarrow anti-phase, not same. False. …

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