When a particle executes $SHM$ the nature of graphical representation of velocity as a function of displacement is :
A
circular
B
elliptical
C
parabolic
D
straight line
JEE MAIN 2021, Medium
Download our app for free and get started
B
elliptical
b For a particle executing SHM,
$x = A \sin (\omega t +\phi)$
$v =\omega A \cos (\omega t +\phi)$
$\Rightarrow \frac{ v ^{2}}{\omega^{2} A ^{2}}+\frac{ x ^{2}}{ A ^{2}}=1 \Rightarrow$ equation of ellipse between $v$ and $x.$
Download our app
and 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.*
An ideal spring with spring-constant $K$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is
The kinetic energy of a particle executing $S.H.M.$ is $16\, J$ when it is in its mean position. If the amplitude of oscillations is $25\, cm$ and the mass of the particle is $5.12\, kg$, the time period of its oscillation is
A particle executes linear simple harmonic motion with an amplitude of $2\, cm$. When the particle is at $1\, cm$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
A particle of mass $0.50 \mathrm{~kg}$ executes simple harmonic motion under force $\mathrm{F}=-50\left(\mathrm{Nm}^{-1}\right) \mathrm{x}$. The time period of oscillation is $\frac{x}{35} s$. The value of $x$ is . . . . .(Given $\pi=\frac{22}{7}$ )
The equation of motion of a particle is $\frac{{{d^2}y}}{{d{t^2}}} + Ky = 0$, where $K$ is positive constant. The time period of the motion is given by
$Assertion :$ In simple harmonic motion, the velocity is maximum when the acceleration is minimum.
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$