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 $v_1$ is given by the derivative of $y_1$ with respect to time $t$: $v_1 = \frac{dy_1}{dt} = 0.1 	imes 100\pi \cos(100\pi t + \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The velocity $v_1$ is given by the derivative of $y_1$ with respect to time $t$: $v_1 = \frac{dy_1}{dt} = 0.1 	imes 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.The velocity $v_2$ is given by the derivative of $y_2$ with respect to time $t$: $v_2 = \frac{dy_2}{dt} = 0.1 	imes (-100\pi) \sin(100\pi t) = -10\pi \sin(100\pi t)$.To compare the phase,we express $v_2$ as a cosine function: $v_2 = 10\pi \cos(100\pi t + \frac{\pi}{2})$.The phase of $v_1$ is $\phi_1 = 100\pi t + \frac{\pi}{3}$ and the phase of $v_2$ is $\phi_2 = 100\pi t + \frac{\pi}{2}$.The phase difference $\Delta\phi = \phi_1 - \phi_2 = (100\pi t + \frac{\pi}{3}) - (100\pi t + \frac{\pi}{2}) = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$.

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

Similar Questions

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

In a $P$-type semiconductor,germanium is doped with:

The specific rotation of the equilibrium mixture of $\alpha-D$-glucose and $\beta-D$-glucose is ....... $^\circ$.

$A$ particle of mass $m$ moves in the $XY$ plane with a velocity $V$ along the straight line $AB$. If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B$,then

At metaphase,chromosomes are attached to the spindle fibres by their

Vedclass Test Series

Exam Paper Generator

Online Exam Module