A particle executes linear simple harmonic motion with an amplitu… - Vidyadip

MCQ

A particle executes linear simple harmonic motion with an amplitude of $2 cm$. When the particle is at $1 cm$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A
$\frac{1}{2 \pi \sqrt{3}}$
B
$2 \pi \sqrt{3}$
✓
$\frac{2 \pi}{\sqrt{3}}$
D
$\frac{\sqrt{3}}{2 \pi}$

✓

Answer

Correct option: C.

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

(c) Velocity $v=\omega \sqrt{A^2-x^2}$ and acceleration $=\omega^2 x$Now given, $\omega^2 x=\omega \sqrt{A^2-x^2} \Rightarrow \omega^2 \cdot 1=\omega \sqrt{2^2-1^2}$ $\Rightarrow \omega=\sqrt{3} \quad \therefore T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{3}}$

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