The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $\frac{v d v}{d x}=-\omega^2 x$ with the initial condition $v=v_0$ at $x=0$. The velocity $v$, when displacement is $x$, is
AIIMS 2015, Diffcult
Download our app for free and get started
(b)
As it is $SHM$ so the equation of motion will be $F=- kx$ or $\frac{ vdv }{ dx }=-\omega^2 x$
Now integrating the expression with boundary condition, $\int \limits_{ v _0}^v v d v=-\omega^2 \int \limits_0^\pi xdx$
or $\frac{1}{2}\left[v^2-v_0^2\right]=-\frac{\omega^2 x^2}{2}$
or $v =\sqrt{ v _0^2-\omega^2 x ^2}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1The $S.H.M.$ of a particle is given by the equations $=2 \sin \omega t+4 \cos \omega t$. Its amplitude of oscillation is ........ unitsView Solution
- 2A particle is executing Simple Harmonic Motion $(SHM)$. The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will beView Solution
- 3A uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$ . In the position shown in figure, the strings are in their natural length. When the stick is displaced through a small angle $\theta $ and released. The stickView Solution
- 4There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4,$ then the new time period of the pendulum isView Solution
- 5A particle of mass $m$ in a unidirectional potential field have potential energy $U(x)=\alpha+2 \beta x^2$, where $\alpha$ and $\beta$ are positive constants. Find its time period of oscillation.View Solution
- 6The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is-View Solution
- 7A simple harmonic oscillator has a period of $0.01 \,sec$ and an amplitude of $0.2\, m$. The magnitude of the velocity in $m{\sec ^{ - 1}}$ at the centre of oscillation isView Solution
- 8A particle executing simple harmonic motion with amplitude of $0.1 \,m$. At a certain instant when its displacement is $0.02 \,m$, its acceleration is $0.5 \,m/s^2$. The maximum velocity of the particle is (in $m/s$)View Solution
- 9A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period $T$. If the mass is increased by m then the time period becomes $\left( {\frac{5}{4}T} \right)$. The ratio of $\frac{m}{{M}}$ isView Solution
- 10The amplitude of the vibrating particle due to superposition of two $SHMs,$View Solution
$y_1 = \sin \left( {\omega t + \frac{\pi }{3}} \right)$ and $y_2 = \sin \omega t$ is :