A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , therefore
The total energy of the system is $\frac{1}{2}K{A^2}$
The frequency is $\frac{1}{{2\pi }}\sqrt {\frac{K}{M}} $
The maximum velocity occurs, when $x = 0$
All the above are correct
A mass of $2.0\, kg$ is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is $200\, N/m.$ What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take $g = 10 m/s^2$).
When a mass $m$ is attached to a spring, it normally extends by $0.2\, m$. The mass $m$ is given a slight addition extension and released, then its time period will be
$Assertion :$ The time-period of pendulum, on a satellite orbiting the earth is infinity.
$Reason :$ Time-period of a pendulum is inversely proportional to $\sqrt g$
Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K _{1}$ and $K _{2}$ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of $A$ and $B$ is
Fill in the blank : Force constant of spring is $0.5\, Nm^{-1}$. The force necessary to increase the length of $10 \,cm$ of spring will be ..........