MHT CET · Physics · 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
- A \(\sqrt{3} \mathrm{~cm}\)
- B \(\sqrt{6} \mathrm{~cm}\)
- C \(2 \sqrt{3} \mathrm{~cm}\)
- D \(2 \sqrt{6} \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{3} \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
At mean position, \(\mathrm{v}_{\max }=\mathrm{A} \omega\)
\(\begin{array}{ll}
\therefore & \omega=\frac{v_{\max }}{A}=\frac{12}{4}=3 \mathrm{rad} / \mathrm{s} \\
& \text { Now, } v=\omega \sqrt{A^2-\mathrm{x}^2} \\
\therefore & \mathrm{v}^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \\
\therefore & x^2=A^2-\frac{v^2}{\omega^2} \\
\therefore & x=\sqrt{16-\frac{36}{9}}=\sqrt{12}=2 \sqrt{3} \mathrm{~cm}
\end{array}\)
\(\begin{array}{ll}
\therefore & \omega=\frac{v_{\max }}{A}=\frac{12}{4}=3 \mathrm{rad} / \mathrm{s} \\
& \text { Now, } v=\omega \sqrt{A^2-\mathrm{x}^2} \\
\therefore & \mathrm{v}^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}^2\right) \\
\therefore & x^2=A^2-\frac{v^2}{\omega^2} \\
\therefore & x=\sqrt{16-\frac{36}{9}}=\sqrt{12}=2 \sqrt{3} \mathrm{~cm}
\end{array}\)
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