When a particle executes SHM,the nature of th | vedclass

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

Question

(B) For a particle executing $SHM$,the displacement is given by $x = A \sin(\omega t + \phi)$.Taking the derivative with respect to time,the velocity

Vedclass · Accepted Answer

elliptical

Vedclass · Answer

(B) For a particle executing $SHM$,the displacement is given by $x = A \sin(\omega t + \phi)$.Taking the derivative with respect to time,the velocity is $v = \frac{dx}{dt} = \omega A \cos(\omega t + \phi)$.From the displacement equation,$\sin(\omega t + \phi) = \frac{x}{A}$.From the velocity equation,$\cos(\omega t + \phi) = \frac{v}{\omega A}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we get:$(\frac{x}{A})^2 + (\frac{v}{\omega A})^2 = 1$.Rearranging this,we get $\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1$.This is the standard equation of an ellipse,$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a = A$ and $b = \omega A$.

When a particle executes $SHM$,the nature of the graphical representation of velocity as a function of displacement is:

Similar Questions

$A$ simple harmonic oscillation is represented by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. Its speed is maximum when $t$ equals

Two $S$.$H$.Ms. are represented by equations $y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$ and $y_2 = 0.1 \cos (100 \pi t)$. The phase difference between the speeds of the two particles is

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

$A$ particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position,the magnitude of its velocity is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

Vedclass Test Series

Exam Paper Generator

Online Exam Module