Two simple harmonic motions are represented b | vedclass

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

Question

(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 	imes 100\pi \cos(100\pi t +

Vedclass · Accepted Answer

$\frac{-\pi}{6}$

Vedclass · Answer

(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 	imes 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.The velocity of the second particle is given by the derivative of its displacement: ${v_2} = \frac{d{y_2}}{dt} = -0.1\pi \sin(\pi t) = 0.1\pi \cos(\pi t + \frac{\pi}{2})$.The phase of the velocity of the first particle is ${\phi_1} = 100\pi t + \frac{\pi}{3}$,and the phase of the velocity of the second particle is ${\phi_2} = \pi t + \frac{\pi}{2}$.However,the question asks for the phase difference between the two velocities. Since the frequencies are different ($100\pi$ vs $\pi$),the phase difference is time-dependent. Assuming the question implies the difference in initial phases at $t=0$:$\Delta \phi = \phi_1(0) - \phi_2(0) = \frac{\pi}{3} - \frac{\pi}{2} = \frac{2\pi - 3\pi}{6} = -\frac{\pi}{6}$.

Two simple harmonic motions are represented by the equations ${y_1} = 0.1 \sin(100\pi t + \frac{\pi}{3})$ and ${y_2} = 0.1 \cos(\pi t)$. The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

Similar Questions

$A$ 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

$A$ body of mass '$m$' performs linear $S$.$H$.$M$. given by the equation $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$. The total energy of the particle at any instant is

Two simple harmonic motions are represented as $y_1 = 10 \sin \omega t$ and $y_2 = 10 \sin \omega t + 5 \cos \omega t$. The ratio of the amplitudes of $y_1$ and $y_2$ is

The displacement of a particle is given by $x = 3\sin(5\pi t) + 4\cos(5\pi t)$. The amplitude of the particle is

Two particles $P$ and $Q$ perform $S.H.M.$ of same amplitude $a$ and frequency along the same straight line. At a particular instant,the maximum distance between the two particles is $\sqrt{2} a$. The initial phase difference between them is

Vedclass Test Series

Exam Paper Generator

Online Exam Module