A particle executing SHM along a straight lin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .

$A$ body of mass $5\, g$ is executing $S.H.M.$ about a point with amplitude $10\, cm$. Its maximum velocity is $100\, cm/s$. Its velocity will be $50\, cm/s$ at a distance of:

The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position,the speed of the particle is $16 \,cm/s$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3} \,cm/s$ will be .... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module