(B) The acceleration of a particle in $\text{S.H.M.}$ is given by $a = -\omega^2 x$. Taking the magnitude,$|a| = \omega^2 |x|$.Given $x = 1 \ cm$ and

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(B) The acceleration of a particle in $\text{S.H.M.}$ is given by $a = -\omega^2 x$. Taking the magnitude,$|a| = \omega^2 |x|$.Given $x = 1 \ cm$ and $a = 3 \ cm s^{-2}$,we have $3 = \omega^2 (1)$,which implies $\omega^2 = 3 \ s^{-2}$.The velocity of a particle in $\text{S.H.M.}$ is given by $v = \omega \sqrt{A^2 - x^2}$.Given $v = 6 \ cm s^{-1}$ at $x = 2 \ cm$,we substitute the values:$6 = \sqrt{3} \sqrt{A^2 - 2^2}$.Squaring both sides:$36 = 3 (A^2 - 4)$.Dividing by $3$:$12 = A^2 - 4$.$A^2 = 16$.$A = 4 \ cm$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

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$A$ particle is executing simple harmonic motion $\text{(S.H.M.).}$ Its acceleration at a distance of $1 \ cm$ from the mean position is $3 \ cm s^{-2}$. If its velocity is $6 \ cm s^{-1}$ when it is at a distance of $2 \ cm$ from its mean position,then the amplitude of $\text{S.H.M.}$ is,

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A
$5 \ cm$
B
$4 \ cm$
C
$2 \sqrt{3} \ cm$
D
$3 \sqrt{2} \ cm$

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Oscillations

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Velocity of Simple Harmonic Motion

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The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

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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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The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

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The velocity of a particle performing simple harmonic motion,when it passes through its mean position is

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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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$A$ particle is executing simple harmonic motion $\text{(S.H.M.).}$ Its acceleration at a distance of $1 \ cm$ from the mean position is $3 \ cm s^{-2}$. If its velocity is $6 \ cm s^{-1}$ when it is at a distance of $2 \ cm$ from its mean position,then the amplitude of $\text{S.H.M.}$ is,

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The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

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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