Two SHM are represented by equations,y_1 = 6c | vedclass

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

Question

(A) Given equations are:$y_1 = 6 \cos \left( 6 \pi t + \frac{\pi}{6} \right)$$y_2 = 3 \left( \sqrt{3} \sin 3 \pi t + \cos 3 \pi t \right)$For $y_1$,th

Vedclass · Accepted Answer

The ratio of their amplitudes is $1$.

Vedclass · Answer

(A) Given equations are:$y_1 = 6 \cos \left( 6 \pi t + \frac{\pi}{6} \right)$$y_2 = 3 \left( \sqrt{3} \sin 3 \pi t + \cos 3 \pi t \right)$For $y_1$,the amplitude $A_1 = 6$.For $y_2$,we can rewrite it as:$y_2 = 6 \left( \frac{\sqrt{3}}{2} \sin 3 \pi t + \frac{1}{2} \cos 3 \pi t \right)$Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \phi = \frac{\sqrt{3}}{2}$ and $\sin \phi = \frac{1}{2}$,which gives $\phi = \frac{\pi}{6}$.$y_2 = 6 \sin \left( 3 \pi t + \frac{\pi}{6} \right)$Thus,the amplitude $A_2 = 6$.The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{6}{6} = 1$.Therefore,option $A$ is the correct answer.

Two $SHM$ are represented by equations,$y_1 = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right)$ and $y_2 = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)$. Which of the following statements is true?

Similar Questions

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be:

The amplitude of the vibrating particle due to the superposition of two $SHMs$,$y_1 = \sin \left( \omega t + \frac{\pi}{3} \right)$ and $y_2 = \sin \omega t$ is:

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by a distance $X_0$ $(X_0 > A)$. If the maximum separation between them is $(X_0 + A)$,the phase difference between their motions is:

Two simple harmonic motions are represented by the equations $y_{1} = 10 \sin(3 \pi t + \frac{\pi}{3})$ and $y_{2} = 5(\sin 3 \pi t + \sqrt{3} \cos 3 \pi t)$. The ratio of the amplitude of $y_{1}$ to $y_{2}$ is $x : 1$. The value of $x$ is ...... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module