(B) The stroke of the piston is equal to twice the amplitude $(2A)$.Given,$2A = 6\,m$,so the amplitude $A = 3\,m$.The angular frequency $\omega = 200\

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(B) The stroke of the piston is equal to twice the amplitude $(2A)$.Given,$2A = 6\,m$,so the amplitude $A = 3\,m$.The angular frequency $\omega = 200\,rad\,min^{-1}$.To convert the angular frequency into $rad\,s^{-1}$,we divide by $60$:$\omega = \frac{200}{60} = \frac{10}{3}\,rad\,s^{-1}$.The maximum speed $(V_{\max})$ of a particle executing simple harmonic motion is given by the formula $V_{\max} = A\omega$.Substituting the values: $V_{\max} = 3 \times \frac{10}{3} = 10\,m\,s^{-1}$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

Medium

The piston in the cylinder head of a locomotive has a stroke of $6\,m$ (which is twice the amplitude). If the piston is executing simple harmonic motion with an angular frequency of $200\,rad\,min^{-1}$,its maximum speed is .... $m\,s^{-1}$.

A
$5$
B
$10$
C
$15$
D
$20$

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Oscillations

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Velocity of Simple Harmonic Motion

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

MediumAIPMT 1991

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In the figure,what will be the sign of the velocity of the point $P_1$,which is the projection of the velocity of the reference particle $P$? $P$ is moving in a circle of radius $R$ in an anti-clockwise direction.

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$A$ particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium position is at $x = 0$,the force constant is $K$,and the amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be:

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The maximum speed of a particle executing $S.H.M.$ is $1 \ m/s$ and its maximum acceleration is $1.57 \ m/s^2$. The time period of the particle will be .... $s$.

Easy

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$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.

EasyAIIMS 2009

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The piston in the cylinder head of a locomotive has a stroke of $6\,m$ (which is twice the amplitude). If the piston is executing simple harmonic motion with an angular frequency of $200\,rad\,min^{-1}$,its maximum speed is .... $m\,s^{-1}$.

Similar Questions

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

In the figure,what will be the sign of the velocity of the point $P_1$,which is the projection of the velocity of the reference particle $P$? $P$ is moving in a circle of radius $R$ in an anti-clockwise direction.

$A$ particle of mass $m$ is performing linear simple harmonic motion. Its equilibrium position is at $x = 0$,the force constant is $K$,and the amplitude of $SHM$ is $A$. The maximum power supplied by the restoring force to the particle during $SHM$ will be:

The maximum speed of a particle executing $S.H.M.$ is $1 \ m/s$ and its maximum acceleration is $1.57 \ m/s^2$. The time period of the particle will be .... $s$.

$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.

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$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.