(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

Medium

$A$ 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 performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions 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

MediumMHT CET 2024

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$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

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$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is . . . . . . .

DifficultJEE MAIN 2024

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$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

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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 $50 \ cm/s$?

Similar Questions

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

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

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is . . . . . . .

$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

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