Question 282: Oscillations

Question

A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle is $12 \mathrm{~cm} / \mathrm{s}$. When the speed of the particle becomes $6 \mathrm{~cm} / \mathrm{s}$, the distance of the particle from mean position is

Accepted Answer

$$\begin{array}{ll} 
& 	ext { At mean position, } \mathrm{v}_{\max }=\mathrm{A} \omega \
	herefore & \omega=\frac{\mathrm{v}_{\max }}{\mathrm{A}}=\frac{12}{4}=3 \mathrm{rad} / \mathrm{s} \
& 	ext { Now, } \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \
	herefore & \mathrm{v}^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}^2ight) \
	herefore & \mathrm{x}^2=\mathrm{A}^2-\frac{\mathrm{v}^2}{\omega^2} \
	herefore \quad & \mathrm{x}=\sqrt{16-\frac{36}{9}}=\sqrt{12}=2 \sqrt{3} \mathrm{~cm}
\end{array}$$

Answer

$\sqrt{3} \mathrm{~cm}$

Answer

$\sqrt{6} \mathrm{~cm}$

Answer

$2 \sqrt{3} \mathrm{~cm}$

Answer

$2 \sqrt{6} \mathrm{~cm}$

Oscillations

A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle is 12 \mathrm{~cm} / \mathrm{s}. When the speed of the particle becomes 6 \mathrm{~cm} / \mathrm{s}, the distance of the particle from mean position is

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