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JEE Main 201419 AprilOnline
Exam
JEE Main
Year
2014
Date
19 April
Mode
Online
Subject
Physics
Chapter
Oscillations

JEE Main 2014 · 19 April · Online · Physics · Oscillations

A body is in simple harmonic motion with time period T = 0.5s and...

A body is in simple harmonic motion with time period T=0.5sT = 0.5\text{s} and amplitude A=1cmA = 1\text{cm}. Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Options

  1. A16cm/s16\text{cm/s}
  2. B6cm/s6\text{cm/s}
  3. C4cm/s4\text{cm/s}
  4. D12cm/s12\text{cm/s}

Correct answer

D. 12cm/s12\text{cm/s}

Step-by-step solution

∴x=Asin\omegat∴ x = Asin \omegat

A2=Asin\omegat\dfrac{A}{2} = Asin \omegat

\omegat=π6\omegat = \dfrac{\pi}{6}

t=π6ωt = \dfrac{\pi}{6\omega}

=π6×2π= \dfrac{\pi}{6 \times 2 \pi}

t=T12t = \dfrac{T}{12}

∴Timetakentoreachfromx=0tox=A2isT12∴ Timetakentoreachfromx = 0tox = \dfrac{A}{2}is\dfrac{T}{12}

∴Averagevelocity=DisplacementTime∴ Averagevelocity = \dfrac{Displacement}{Time}

=A/2T/12= \dfrac{A / 2}{T / 12}

=AT×6= \dfrac{A}{T} \times 6

=6×10.5= \dfrac{6 \times 1}{0.5}

=12cm/s= 12cm / s
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