Three simple harmonic motions of equal amplit | vedclass

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

Question

(B) Let the three SHMs be represented by phasors of magnitude $A$ at angles $0^{\circ}$,$60^{\circ}$,and $120^{\circ}$ with respect to the first motio

Vedclass · Accepted Answer

$2\,A$

Vedclass · Answer

(B) Let the three SHMs be represented by phasors of magnitude $A$ at angles $0^{\circ}$,$60^{\circ}$,and $120^{\circ}$ with respect to the first motion.The resultant amplitude $A_{	ext{net}}$ is given by the vector sum of these three phasors:$A_{	ext{net}} = \sqrt{(A + A\cos 60^{\circ} + A\cos 120^{\circ})^2 + (0 + A\sin 60^{\circ} + A\sin 120^{\circ})^2}$$A_{	ext{net}} = \sqrt{(A + A/2 - A/2)^2 + (0 + A\sqrt{3}/2 + A\sqrt{3}/2)^2}$$A_{	ext{net}} = \sqrt{A^2 + (A\sqrt{3})^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A$Alternatively,using the phasor diagram,the resultant of the first and third phasors (which are at $120^{\circ}$ to each other) is a vector of magnitude $A$ directed along the same line as the second phasor. Adding this to the second phasor gives $A + A = 2A$.

Three simple harmonic motions of equal amplitudes $A$ and equal time periods in the same direction combine. The phase of the second motion is $60^{\circ}$ ahead of the first and the phase of the third motion is $60^{\circ}$ ahead of the second. Find the amplitude of the resultant motion.

Similar Questions

$A$ 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

The equation of $SHM$ is given as:
$x = 3 \sin(20\pi t) + 4 \cos(20\pi t)$,
where $x$ is in $cm$ and $t$ is in $seconds$. The amplitude is ..... $cm$.

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

$A$ particle is subjected to two simple harmonic motions as:
$x_1 = \sqrt{7} \sin(5t) \ cm$
and $x_2 = 2\sqrt{7} \sin(5t + \frac{\pi}{3}) \ cm$
where $x$ is displacement and $t$ is time in seconds.
The maximum acceleration of the particle is $x \times 10^{-2} \ ms^{-2}$. The value of $x$ is

Two particles $p$ and $q$ perform $SHM$ with the same amplitude $a$ and the same frequency $f$ along a straight line. The maximum distance between the two particles is $a\sqrt{2}$. The initial phase difference between the particles is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module