$A$ particle executes two types of $SHM$. $x_1 = A_1 \sin \omega t$ and $x_2 = A_2 \sin [\omega t + \frac{\pi}{3}]$.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.
(N/A) The resultant displacement $x = x_1 + x_2$. At $t = 0$,$x_1 = A_1 \sin(0) = 0$ and $x_2 = A_2 \sin(\frac{\pi}{3}) = A_2 \frac{\sqrt{3}}{2}$. Thus,the displacement is $x = \frac{\sqrt{3}}{2} A_2$.
$(b)$ The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\frac{\pi}{3})}$. Since $\cos(\frac{\pi}{3}) = 0.5$,$A = \sqrt{A_1^2 + A_2^2 + A_1 A_2}$. The maximum speed is $v_{\max} = A \omega = \omega \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(c)$ The maximum acceleration is $a_{\max} = A \omega^2 = \omega^2 \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(b)$ The resultant amplitude $A$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\frac{\pi}{3})}$. Since $\cos(\frac{\pi}{3}) = 0.5$,$A = \sqrt{A_1^2 + A_2^2 + A_1 A_2}$. The maximum speed is $v_{\max} = A \omega = \omega \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
$(c)$ The maximum acceleration is $a_{\max} = A \omega^2 = \omega^2 \sqrt{A_1^2 + A_2^2 + A_1 A_2}$.
Explore More
Similar Questions
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T$,respectively. At time $t=0$,one particle has displacement $A$ while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time $t$,then $t$ is
DifficultJEE MAIN 2016
View SolutionThe resultant of two rectangular simple harmonic motions of the same frequency and equal amplitudes but differing in phase by $\frac{\pi}{2}$ is
MediumAIPMT 1997
View SolutionTwo simple harmonic motions are represented by $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$. The ratio of their amplitudes is
$A$ vibratory motion is represented by $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$. The resultant amplitude of the motion is
Medium
View SolutionTwo mutually perpendicular simple harmonic vibrations have the same amplitude,frequency,and phase. When they superimpose,the resultant form of vibration will be:
Medium
View SolutionVedclass 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