The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.
(N/A) In the displacement-time graph of $SHM$,the crests and troughs represent the points of maximum displacement,where the velocity of the oscillator is zero.
$(i)$ At points $A$,$C$,$E$,and $G$,the displacement is at its extreme value,so the velocity is zero.
$(ii)$ At the mean position (where displacement is zero),the speed of the oscillator is maximum. Hence,at points $B$,$D$,$F$,and $H$,the speed is maximum.
$(i)$ At points $A$,$C$,$E$,and $G$,the displacement is at its extreme value,so the velocity is zero.
$(ii)$ At the mean position (where displacement is zero),the speed of the oscillator is maximum. Hence,at points $B$,$D$,$F$,and $H$,the speed is maximum.
Explore More
Similar Questions
$A$ particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$,and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$,then
Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is
DifficultAIPMT 2011
View SolutionWhat is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?
$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:
$A$ particle performs simple harmonic motion with a period of $3 \ s$. The time taken by it to cover a distance equal to half the amplitude from the mean position is $\left[\sin 30^{\circ}=0.5\right]$.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo