MCQ
A particle starts simple harmonic motion from the mean position. Its amplitude is $a$ and total energy $E$. At one instant its kinetic energy is $3E/4.$ Its displacement at that instant is
  • A
    ${y}={a}$
  • $y=\frac{a}{2}$
  • C
    $y=\frac{a \sqrt{3}}{2}$
  • D
    $y=\frac{a}{\sqrt{2}}$

Answer

Correct option: B.
$y=\frac{a}{2}$
b
(b) $\frac{K}{E} = \frac{{\frac{1}{2}m{\omega ^2}({a^2} - {y^2})}}{{\frac{1}{2}m{\omega ^2}{a^2}}}$

$= \frac{{{a^2} - {y^2}}}{{{a^2}}} $

$= 1 - \frac{{{y^2}}}{{{a^2}}}$

So, $\frac{{\left( {\frac{{3E}}{4}} \right)}}{E} = 1 - \frac{{{y^2}}}{{{a^2}}}$

$\Rightarrow \frac{{{y^2}}}{{{a^2}}} = 1 - \frac{3}{4} = \frac{1}{4}$

$ \Rightarrow y = \frac{a}{2}$.

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