A particle executes linear S.H.M. with amplit | vedclass

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

Question

(B) For a particle in $S.H.M.$, the velocity $v$ at position $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The acceleration $a$ at position $x$ is giv

Vedclass · Accepted Answer

$\frac{6\pi}{\sqrt{7}} 	ext{ s}$

Vedclass · Answer

(B) For a particle in $S.H.M.$, the velocity $v$ at position $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The acceleration $a$ at position $x$ is given by $a = \omega^2 x$.Given that the magnitude of velocity equals the magnitude of acceleration: $|v| = |a|$.$\omega \sqrt{A^2 - x^2} = \omega^2 x$.$\sqrt{A^2 - x^2} = \omega x$.Given $A = 4 	ext{ cm}$ and $x = 3 	ext{ cm}$:$\sqrt{4^2 - 3^2} = \omega (3)$.$\sqrt{16 - 9} = 3\omega$.$\sqrt{7} = 3\omega \implies \omega = \frac{\sqrt{7}}{3} 	ext{ rad/s}$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{(\sqrt{7}/3)} = \frac{6\pi}{\sqrt{7}} 	ext{ s}$.

$A$ particle executes linear $S.H.M.$ with amplitude $A = 4 \text{ cm}$. The magnitude of velocity and acceleration is equal when it is at a distance $x = 3 \text{ cm}$ from the mean position. The time period of the oscillation is:

Similar Questions

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $16 \ s$. After $2 \ s$,its velocity is $\pi \ m \ s^{-1}$. The amplitude of oscillation is (Given: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$).

The velocity of a particle performing simple harmonic motion,when it passes through its mean position is

For a particle executing simple harmonic motion $(SHM)$,at its mean position

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is

The kinetic energy of a particle executing $S.H.M.$ is $16 \, J$ when it is at its mean position. If the mass of the particle is $0.32 \, kg$,then what is the maximum velocity of the particle in $m/s$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module