A particle executes simple harmonic motion wi | vedclass

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

Question

(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.

Vedclass · Accepted Answer

$\frac{8\pi}{3}$

Vedclass · Answer

(C) Given,amplitude $A = 5\, cm$ and displacement $x = 4\, cm$.The magnitude of velocity in simple harmonic motion is $|v| = \omega \sqrt{A^2 - x^2}$.The magnitude of acceleration in simple harmonic motion is $|a| = \omega^2 x$.According to the problem,$|v| = |a|$ at $x = 4\, cm$:$\omega \sqrt{A^2 - x^2} = \omega^2 x$Substituting the values:$\omega \sqrt{5^2 - 4^2} = \omega^2 	imes 4$$\omega \sqrt{25 - 16} = 4\omega^2$$\omega 	imes 3 = 4\omega^2$$\omega = \frac{3}{4}\, rad/s$The periodic time $T$ is given by $T = \frac{2\pi}{\omega}$:$T = \frac{2\pi}{3/4} = \frac{8\pi}{3}\, s$.

$A$ 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

Similar Questions

$A$ particle executes linear simple harmonic motion with an amplitude of $3\,cm$. When the particle is at $2\,cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .

The velocity-time diagram of a harmonic oscillator is shown in the figure. The frequency of oscillation is ..... $Hz$.

$A$ particle is performing $S.H.M.$ about its mean position with an amplitude $a$ and periodic time $T$. The speed of the particle when its displacement from the mean position is $\frac{a}{3}$ will be:

$A$ particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period will the velocity of the particle be equal to half of its maximum velocity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module