A particle executes $S.H.M.,$ the graph of velocity as a function of displacement is :-
JEE MAIN 2021, Medium
Download our app for free and get started
$v ^{2}=\omega^{2}\left( A ^{2}- x ^{2}\right)$
$\frac{ v ^{2}}{\omega^{2}}+ x ^{2}= A ^{2}$
$\frac{ v ^{2}}{(\omega A )^{2}}+\frac{ x ^{2}}{ A ^{2}}=1$
This is an equation of an ellipse.
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
- 1A second's pendulum is placed in a space laboratory orbiting around the earth at a height $3R$, where $R$ is the radius of the earth. The time period of the pendulum isView Solution
- 2A particle executes simple harmonic motion represented by displacement function as $x(t)=A \sin (\omega t+\phi)$View Solution
If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .
- 3The function $(sin\,\omega t -cos\,\omega t)$ representsView Solution
- 4A vibratory motion is represented by $x = 2\,A\,\cos \,\omega t + A\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right) + A\,\cos \,\left( {\omega t + \pi } \right) + \frac{A}{2}\,\cos \,\left( {\omega t + \frac{{3\pi }}{2}} \right)$ The resultant amplitude of motion isView Solution
- 5If a spring extends by $x$ on loading, then energy stored by the spring is (if $T$ is the tension in the spring and $K$ is the spring constant)View Solution
- 6A plank with a small block on top of it is under going vertical $SHM$ . Its period is $2\ sec$ . The minimum amplitude at which the block will separate from plank isView Solution
- 7Two springs of force constants $300\, N/m$ (Spring $A$) and $400\, N/m$ (Spring $B$) are joined together in series . The combination is compressed by $8.75\, cm$. The ratio of energy stored in $A$ and $B$ is $\frac{{{E_A}}}{{{E_B}}}$. Then $\frac{{{E_A}}}{{{E_B}}}$ is equal toView Solution
- 8A simple pendulum consisting of a ball of mass $m$ tied to a thread of length $l$ is made to swing on a circular arc of angle $\theta $ in a vertical plane. At the end of this arc, another ball of mass $m$ is placed at rest. The momentum transferred to this ball at rest by the swinging ball isView Solution
- 9The bob of a simple pendulum of mass m and total energy $E$ will have maximum linear momentum equal toView Solution
- 10When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ isView Solution