The velocity at the mean position of a partic | vedclass

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

Question

(C) The velocity of a particle in $S.H.M.$ at a displacement $x$ is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}v$

Vedclass · Answer

(C) The velocity of a particle in $S.H.M.$ at a displacement $x$ is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.At the mean position,$x = 0$,so the velocity $v = \omega \sqrt{a^2 - 0^2} = a\omega$.At a distance equal to half of the amplitude,$x = \frac{a}{2}$.Substituting this into the velocity formula: $V' = \omega \sqrt{a^2 - (\frac{a}{2})^2} = \omega \sqrt{a^2 - \frac{a^2}{4}} = \omega \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}}{2}a\omega$.Since $v = a\omega$,we have $V' = \frac{\sqrt{3}}{2}v$.

The velocity at the mean position of a particle executing $S.H.M.$ is $v$. What is the velocity of the particle at a distance equal to half of the amplitude?

Similar Questions

The displacement of a particle executing $SHM$ is given by $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$. The time period of motion is $3.14 \ s$. The velocity of the particle at $t = 0$ is . . . . . . $m/s$.

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

Obtain the velocity of a particle executing simple harmonic motion $(SHM)$ by considering the projection of a particle undergoing uniform circular motion.

Vedclass Test Series

Exam Paper Generator

Online Exam Module