(C) Given: Amplitude $a = 10 \ cm$,Maximum velocity $v_{\max} = 100 \ cm/s$.We know that $v_{\max} = a\omega$,so $100 = 10 \times \omega$,which gives

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(C) Given: Amplitude $a = 10 \ cm$,Maximum velocity $v_{\max} = 100 \ cm/s$.We know that $v_{\max} = a\omega$,so $100 = 10 \times \omega$,which gives $\omega = 10 \ rad/s$.The velocity $v$ at a displacement $y$ is given by the formula $v = \omega \sqrt{a^2 - y^2}$.Substituting the given values: $50 = 10 \sqrt{10^2 - y^2}$.Dividing by $10$: $5 = \sqrt{100 - y^2}$.Squaring both sides: $25 = 100 - y^2$.$y^2 = 100 - 25 = 75$.$y = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \ cm$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

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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 $50 \ cm/s$?

A
$5 \ cm$
B
$5\sqrt{2} \ cm$
C
$5\sqrt{3} \ cm$
D
$10\sqrt{2} \ cm$

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Oscillations

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Velocity of Simple Harmonic Motion

$A$ particle executes $SHM$. Its velocities are $v_1$ and $v_2$ at displacements $x_1$ and $x_2$ from the mean position,respectively. The frequency of oscillation will be

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$A$ particle is executing $S.H.M.$ If its amplitude is $2 \, m$ and periodic time is $2 \, s$,then the maximum velocity of the particle will be:

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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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