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13. oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics13. oscillations1 Mark
MCQ
A particle executes simple harmonic motion with an amplitude of $4 \,cm$. At the mean position the velocity of the particle is $10\, cm/s$. The distance of the particle from the mean position when its speed becomes $5 \,cm/s$ is
  • A
    $\sqrt 3 \,cm$
  • B
    $\sqrt 5 \,cm$
  • ✓
    $2(\sqrt 3 )\,cm$
  • D
    $2(\sqrt 5 )\,cm$
✓

Answer

Correct option: C.
$2(\sqrt 3 )\,cm$
c
(c) ${v_{\max }} = a\omega $

==> $\omega = \frac{{{v_{\max }}}}{a} = \frac{{10}}{4}$ 

Now, $v = \omega \sqrt {{a^2} - {y^2}} $

==> ${v^2} = {\omega ^2}({a^2} - {y^2})$

==> ${y^2} = {a^2} - \frac{{{v^2}}}{{{\omega ^2}}}$ 

$y = \sqrt{{a^2} - \frac{{{v^2}}}{{{\omega ^2}}}}$

$= \sqrt{{4^2} - \frac{5^2}{{({\frac{10}{4}})^2}}}$

$ = 2\sqrt 3 \,cm$

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