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A Particle executes simple harmonic motion with an amplitude of 9 cm. When the Particle is at 6 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its Periodic time
Option: 1
$T=\frac{7 \pi}{\sqrt{5}}$

Option: 2
$T=\frac{2 \pi}{\sqrt{5}}$

Option: 3
$T=\frac{4 \pi}{\sqrt{5}}$

Option: 4
$T=\frac{2 \pi}{\sqrt{5}}$

Answers (1)

$a=-w^2 x \\$

$V_{\text {max }}=A w \\$

$V=w \sqrt{A^2-x^2} \\$

$w^2 \cdot 6=w \sqrt{9^2-6^2} \\$

$6 w=\sqrt{81-36} \\$

$6 w=\sqrt{45} \\$

$6 w=\sqrt{9 \times 5} \\$

$6 w=3 \sqrt{5} \\$

$w=\frac{3 \sqrt{5}}{6} \\$

$w=\frac{\sqrt{5}}{2} \\$ $[\because w=\frac{2 \pi}{T}]$

$\frac{2 \pi}{T}=\frac{\sqrt{5}}{2} \\$

$4 \pi=\sqrt{5} T \\$

$T=\frac{4 \pi}{\sqrt{5}}$

Posted by

Sumit Saini

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