(C) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.Maximum velocity is $V_{\max} = \omega A$.Giv

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(C) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.Maximum velocity is $V_{\max} = \omega A$.Given that the speed $V = \frac{V_{\max}}{3} = \frac{\omega A}{3}$.Equating the two expressions: $\frac{\omega A}{3} = \omega \sqrt{A^2 - x^2}$.Squaring both sides: $\frac{A^2}{9} = A^2 - x^2$.Rearranging for $x^2$: $x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$.Taking the square root: $x = \sqrt{\frac{8}{9} A^2} = \frac{2\sqrt{2}}{3} A$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

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For a particle executing $S.H.M.$ having amplitude $A$,the speed of the particle is $\left(\frac{1}{3}\right)^{rd}$ of its maximum speed when the displacement from the mean position is

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A
$\frac{3 A}{\sqrt{2}}$
B
$\frac{2 A}{3}$
C
$\frac{2 \sqrt{2}}{3} A$
D
$\frac{\sqrt{2}}{3} A$

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Oscillations

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Velocity of Simple Harmonic Motion

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$A$ particle performing $SHM$ is found at its equilibrium position at $t = 1 \, s$. It is found to have a speed of $0.25 \, m/s$ at $t = 2 \, s$. If the period of oscillation is $6 \, s$,calculate the amplitude of oscillation.

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$A$ particle of mass $5 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$. Its maximum velocity is $100 \ cm/s$. At what distance from the mean position will its velocity be $50 \ cm/s$?

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$A$ particle is making simple harmonic motion along the $x$-axis. If at a distance $x_{1}$ and $x_{2}$ from the mean position the velocities of the particle are $v_{1}$ and $v_{2}$ respectively,the time period of its oscillation is given as -

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For a particle executing $S.H.M.$ having amplitude $A$,the speed of the particle is $\left(\frac{1}{3}\right)^{rd}$ of its maximum speed when the displacement from the mean position is

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