A particle is performing a linear simple harm | vedclass

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

Question

(A) In linear simple harmonic motion $(SHM)$,the velocity $v$ of a particle at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The magnitud

Vedclass · Accepted Answer

$\frac{2 \pi}{\sqrt{3}} \ s$

Vedclass · Answer

(A) In linear simple harmonic motion $(SHM)$,the velocity $v$ of a particle at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The magnitude of acceleration $a$ is given by $a = \omega^2 x$.Given that the particle is midway between the mean position $(x=0)$ and the extreme position $(x=A)$,we have $x = \frac{A}{2}$.According to the problem,the magnitudes of velocity and acceleration are equal,so $v = a$.Substituting the expressions: $\omega \sqrt{A^2 - x^2} = \omega^2 x$.Substituting $x = \frac{A}{2}$: $\omega \sqrt{A^2 - (\frac{A}{2})^2} = \omega^2 (\frac{A}{2})$.$\omega \sqrt{A^2 - \frac{A^2}{4}} = \omega^2 \frac{A}{2} \Rightarrow \omega \sqrt{\frac{3A^2}{4}} = \omega^2 \frac{A}{2}$.$\omega \frac{\sqrt{3}A}{2} = \omega^2 \frac{A}{2}$.Dividing both sides by $\frac{\omega A}{2}$,we get $\omega = \sqrt{3}$.Since the periodic time $T = \frac{2\pi}{\omega}$,we have $T = \frac{2\pi}{\sqrt{3}} \ s$.

$A$ particle is performing a linear simple harmonic motion of amplitude $A$. When it is midway between its mean and extreme position,the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

Similar Questions

$A$ cylindrical log of wood of height $h$ and area of cross-section $A$ floats in water. It is pressed and then released. Show that the log would execute $SHM$ with a time period $T = 2\pi \sqrt{\frac{m}{A\rho g}}$,where $m$ is the mass of the body and $\rho$ is the density of the liquid.

$A$ ring of diameter $2 \ m$ oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre moves in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is .... $m$

$A$ body performs $S.H.M.$ under the action of force $F_1$ with period $T_1$ seconds. If the force is changed to $F_2$,it performs $S.H.M.$ with period $T_2$ seconds. If both forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,the period in seconds will be

$A$ uniform rod of mass $m$ and length $l$ is suspended about its end. The time period of small angular oscillations is ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module