$A$ wire of length $L$,cross-sectional area $A$,and Young's modulus $Y$ is suspended,and a spring of force constant $k$ is attached to its lower end. If a mass $m$ is suspended from the spring and set into oscillations,what will be the time period of the system?

$A$ spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support. Assuming the spring to be made of a uniform wire,the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is

$A$ mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A$. Its speed as it passes through the equilibrium position is $V$. If extended $2A$ and released,the speed of the mass passing through the equilibrium position will be

$A$ mass on a vertical spring begins its motion at rest at $y = 0 \ cm$. It reaches a maximum height of $y = 10 \ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy $(KE)$ versus position is given below. Net force on the mass varies with $y$ as

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

The force-deformation equation for a nonlinear spring fixed at one end is $F = 4x^{1/2}$,where $F$ is the force (expressed in newtons) applied at the other end and $x$ is the deformation expressed in meters.