A particle execute S.H.M. along a straight line. The amplitude of… - Vidyadip

MCQ

A particle execute $S.H.M.$ along a straight line. The amplitude of oscillation is $2 \,cm$. When displacement of particle from the mean position is $1 \,cm$, the magnitude of its acceleration is equal to magnitude of its velocity. The time period of oscillation is ........

A
$\frac{2 \pi}{\sqrt{2}}$
B
$\frac{\sqrt{2}}{2 \pi}$
✓
$\frac{2 \pi}{\sqrt{3}}$
D
$\frac{\sqrt{3}}{2 \pi}$

✓

Answer

Correct option: C.

$\frac{2 \pi}{\sqrt{3}}$

c
(c)

$A=2 \,cm =2 \times 10^{-2}$

$a=-\omega^2 x$

and $v=\omega \sqrt{A^2-x^2}$

$\omega^2 \times 1 \times 10^{-2}=\omega \sqrt{(4-1) 10^{-4}}$

$\omega \times 1 \times 10^{-2}=\sqrt{3} \times 10^{-2}$

$\omega=\sqrt{3}$

$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{3}}$

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