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NCERT Exemplar · Q19

Q.The displacement–time graph of a particle executing S.H.M. is a sinusoidal curve on which eight points are marked in time order: A, B, C, D, E, F, G, H. The points A, D, F and H lie on the time axis (zero displacement, i.e. the mean position). Point B lies at a trough (the most negative displacement) and points C, E and G lie at crests (the most positive displacement); these four points — B, C, E, G — are the extreme positions where the displacement has its maximum magnitude. Identify the marked points at which

(i) the velocity of the oscillator is zero, and
(ii) the speed of the oscillator is maximum.
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A particle in SHM momentarily stops at the extreme positions, so its velocity is zero at the crests and troughs; it moves fastest as it passes through the mean position. On this graph the extremes are the marked points B (trough) and C, E, G (crests), while A, D, F, H sit on the time axis at zero displacement.

Concept

For x=Asin⁡(ωt+ϕ)x=A\sin(\omega t+\phi), the velocity is v=Aωcos⁡(ωt+ϕ)v=A\omega\cos(\omega t+\phi).

  • At an extreme (∣x∣=A|x|=A, crest or trough), cos⁡(⋅)=0⇒v=0\cos(\cdot)=0\Rightarrow v=0.
  • At the mean position (x=0x=0), ∣cos⁡(⋅)∣=1⇒|\cos(\cdot)|=1\Rightarrow speed is maximum, ∣v∣=Aω|v|=A\omega.

Identify the points

  • Extreme positions (velocity =0=0): the trough B and the crests C, E, G — i.e. B, C, E and G. …

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