(D) The angular frequency of the piston is given as $\omega = 200 \; rad/min$.The stroke is defined as twice the amplitude $(2A = 1.0 \; m)$.Therefore

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(D) The angular frequency of the piston is given as $\omega = 200 \; rad/min$.The stroke is defined as twice the amplitude $(2A = 1.0 \; m)$.Therefore,the amplitude $A$ is $A = \frac{1.0 \; m}{2} = 0.5 \; m$.The maximum speed $(v_{\max})$ of a particle executing simple harmonic motion is given by the formula $v_{\max} = A \omega$.Substituting the values,we get $v_{\max} = 0.5 \; m \times 200 \; rad/min = 100 \; m/min$.

Class 11 Physics Oscillations Velocity of Simple Harmonic Motion

Easy

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of $1.0 \; m$. If the piston moves with simple harmonic motion with an angular frequency of $200 \; rad/min$,what is its maximum speed in $m/min$?

A
$136$
B
$80$
C
$125$
D
$100$

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Velocity of Simple Harmonic Motion

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The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of $1.0 \; m$. If the piston moves with simple harmonic motion with an angular frequency of $200 \; rad/min$,what is its maximum speed in $m/min$?

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