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JEE Main 2014 (19 Apr, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2014 (19 Apr, Shift 1)

Q66JEE Main 2014 · 19 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

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Answer: (c)

  1. Mean to A2\frac{A}{2} takes t=T12=0.512=124 st = \frac{T}{12} = \frac{0.5}{12} = \frac{1}{24}\text{ s}
  2. Displacement =A2=0.5 cm= \frac{A}{2} = 0.5\text{ cm}
  3. vavg=0.51/24=12 cm/sv_{avg} = \frac{0.5}{1/24} = 12\text{ cm/s}
  4. So the answer is (c) 12 cm/s.
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