MHT CET · Physics · Oscillations
For a particle executing S.H.M. having amplitude A, the speed of the article is \(\left(\frac{1}{3}\right)^{\text {rd }}\) of its maximum speed when the displacement from the mean position is
- A \(\frac{3 \mathrm{~A}}{\sqrt{2}}\)
- B \(\frac{2 \mathrm{~A}}{3}\)
- C \(\frac{2 \sqrt{2}}{3} \mathrm{~A}\)
- D \(\frac{\sqrt{2}}{3} \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \sqrt{2}}{3} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& V=\omega \sqrt{A^2-x^2}\\
& V_{\max }=\omega A \\
& V=\frac{V_{\max }}{3}=\frac{\omega a}{3} \\
& \frac{\omega A}{3}=\omega \sqrt{A^2-x^2} \\
& \frac{A^2}{9}=\left(A^2-x^2\right) \\
& x^2=A^2-\frac{A^2}{9} \\
& x=\sqrt{\frac{8}{9} A^2} \\
& x=\frac{2 \sqrt{2}}{3} \cdot A
\end{aligned}\)
& V=\omega \sqrt{A^2-x^2}\\
& V_{\max }=\omega A \\
& V=\frac{V_{\max }}{3}=\frac{\omega a}{3} \\
& \frac{\omega A}{3}=\omega \sqrt{A^2-x^2} \\
& \frac{A^2}{9}=\left(A^2-x^2\right) \\
& x^2=A^2-\frac{A^2}{9} \\
& x=\sqrt{\frac{8}{9} A^2} \\
& x=\frac{2 \sqrt{2}}{3} \cdot A
\end{aligned}\)
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