A particle executing SHM along a straight lin | vedclass

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(B) The points $A$ and $B$ are the extreme positions of the $SHM$ because the velocity is zero at these points.The distance between the extreme positi

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$\pi\left(\frac{b-a}{v}\right)$

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(B) The points $A$ and $B$ are the extreme positions of the $SHM$ because the velocity is zero at these points.The distance between the extreme positions $A$ and $B$ is $L = b - a$.The amplitude $A_{amp}$ of the $SHM$ is half the distance between the extreme positions:$A_{amp} = \frac{b - a}{2}$The equilibrium position (mean position) is at the midpoint of $A$ and $B$,which is at a distance of $\frac{a + b}{2}$ from $O$.The velocity of a particle in $SHM$ at a displacement $x$ from the mean position is given by $v = \omega \sqrt{A_{amp}^2 - x^2}$.At the midpoint between $A$ and $B$,the particle is at the mean position,so the displacement $x = 0$.Thus,the velocity at the midpoint is the maximum velocity $v_{max} = \omega A_{amp}$.Given $v_{max} = v$,we have $v = \omega \left(\frac{b - a}{2}\right)$.Solving for angular frequency $\omega$:$\omega = \frac{2v}{b - a}$The time period $T$ is given by:$T = \frac{2\pi}{\omega} = \frac{2\pi}{\left(\frac{2v}{b - a}\right)} = \frac{\pi(b - a)}{v}$

$A$ particle executing $SHM$ along a straight line has zero velocity at points $A$ and $B$ whose distances from $O$ on the same line $OAB$ are $a$ and $b$,respectively. If the velocity at the midpoint between $A$ and $B$ is $v$,then its time period is

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