(D) Given, amplitude, $A = 0.2 \,cm = 2 \times 10^{-3} \,m$. Velocity at the mean position in simple harmonic motion is the maximum velocity, $v_{\tex

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(D) Given, amplitude, $A = 0.2 \,cm = 2 \times 10^{-3} \,m$. Velocity at the mean position in simple harmonic motion is the maximum velocity, $v_{\text{max}} = 5 \,m/s$. We know that the maximum velocity is given by the formula, $v_{\text{max}} = A \omega$. Therefore, the angular frequency $\omega$ is given by, $\omega = \frac{v_{\text{max}}}{A}$. Substituting the values, $\omega = \frac{5}{2 \times 10^{-3}} = 2.5 \times 10^3 = 2500 \,rad/s$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

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$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation is

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A
$1000 \,rad/s$
B
$1500 \,rad/s$
C
$2000 \,rad/s$
D
$2500 \,rad/s$

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$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation is

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