(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,

TS EAMCET 2021 All TS EAMCET

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 amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

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An object is executing simple harmonic motion with an angular frequency $\omega$. If the maximum velocity is $v_{\max}$,then the maximum acceleration of the object 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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$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is

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