$A$ 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
The displacement $y$ of a particle executing periodic motion is given by $y = 4\cos^2(t/2)\sin(1000t)$. This expression may be considered to be a result of the superposition of $..........$ independent harmonic motions.
DifficultIIT 1992
View SolutionTwo particles executing simple harmonic motion as described by $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)$ and $y_2=10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t)$ have amplitudes $A_1$ and $A_2$ respectively. The ratio $A_1: A_2$ is
On the superposition of two harmonic oscillations represented by $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$,a resulting oscillation with the same time period and amplitude is obtained. The value of $\phi_1 - \phi_2$ is .... $^o$
Medium
View SolutionTwo particles execute $SHM$ of same amplitude of $20 \, cm$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $20 \, cm$. Their phase difference in radians is
Consider two SHMs along the same straight line $x_1=A_1 \sin \left(\omega t+\phi_1\right)$ and $x_2=A_2 \sin \left(\omega t+\phi_2\right)$,where $A_1$ and $A_2$ are their amplitudes and $\phi_1$ and $\phi_2$ are their initial phase angles. If $R$ is the resultant amplitude,match the conditions in Column-$I$ with the resultant amplitudes in Column-$II$:
Column-$I$ Column-$II$ $A$. $A_1=A_2=A, \delta=0$ $I$. $A_1+A_2$ $B$. $A_1 \neq A_2, \delta=0$ $II$. $0$ $C$. $A_1=A_2=A, \delta=90^{\circ}$ $III$. $2A$ $D$. $A_1=A_2=A, \delta=180^{\circ}$ $IV$. $A\sqrt{2}$
| Column-$I$ | Column-$II$ |
| $A$. $A_1=A_2=A, \delta=0$ | $I$. $A_1+A_2$ |
| $B$. $A_1 \neq A_2, \delta=0$ | $II$. $0$ |
| $C$. $A_1=A_2=A, \delta=90^{\circ}$ | $III$. $2A$ |
| $D$. $A_1=A_2=A, \delta=180^{\circ}$ | $IV$. $A\sqrt{2}$ |
DifficultMHT CET 2022
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