A particle executes linear simple harmonic motion with an ampilitude of $3\,cm$ . When the particle is at $2\,cm$ from the mean position , the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
NEET 2017, Diffcult
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$\text { Given, } A=3 \mathrm{cm}, x=2 \mathrm{cm}$
The velocity of a particle in simple harmonic motion is given as
$v=\omega \sqrt{A^{2}-x^{2}}$
and magnitude of its acceleration is
$a=\omega^{2} x$
Given $|v|=|a|$
$\therefore \omega \sqrt{A^{2}-x^{2}}-\omega^{2} x$
${\omega x=\sqrt{A^{2}-x^{2}} \text { or } \omega^{2} x^{2}=A^{2}-x^{2}}$
${\omega^{2}=\frac{A^{2}-x^{2}}{x^{2}}=\frac{9-4}{4}=\frac{5}{4}}$
${\omega=\frac{\sqrt{5}}{2}}$
Time period, $T=\frac{2 \pi}{\omega}=2 \pi \cdot \frac{2}{\sqrt{5}}=\frac{4 \pi}{\sqrt{5}} \mathrm{s}$
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