(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.

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(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.The magnitude of acceleration in simple harmonic motion is $|a| = \omega^2 x$.According to the problem,$|v| = |a|$ at $x = 4\, cm$:$\omega \sqrt{A^2 - x^2} = \omega^2 x$Substituting the values:$\omega \sqrt{5^2 - 4^2} = \omega^2 \times 4$$\omega \sqrt{25 - 16} = 4\omega^2$$\omega \times 3 = 4\omega^2$$\omega = \frac{3}{4}\, rad/s$The periodic time $T$ is given by $T = \frac{2\pi}{\omega}$:$T = \frac{2\pi}{3/4} = \frac{8\pi}{3}\, s$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

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$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

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A
$\frac{4\pi}{3}$
B
$\frac{3}{8}\pi$
C
$\frac{8\pi}{3}$
D
$\frac{7}{3}\pi$

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Velocity of Simple Harmonic Motion

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The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is

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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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Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$, respectively. The ratio of the velocities of $P$ and $Q$ when they meet is

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The displacement of a particle executing $SHM$ is given by $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$,where $x$ is in $m$ and $t$ is in $s$. The amplitude and maximum speed of the particle are:

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$A$ particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F = (-25\,x)\,N$. The particle attains a maximum speed of $4\,m/s$ during its oscillation. The amplitude of the motion is $...........\,cm$.

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$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

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The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is

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Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$, respectively. The ratio of the velocities of $P$ and $Q$ when they meet is

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$A$ particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F = (-25\,x)\,N$. The particle attains a maximum speed of $4\,m/s$ during its oscillation. The amplitude of the motion is $...........\,cm$.

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