A force of $20\,dyne$ applied to the end of spring increase its length of $1\, mm$, then force constant will be what ?

When a mass $m$ is hung from the lower end of a spring of neglibgible mass, an extension $x$ is produced in the spring. The time period of oscillation is

One-forth length of a spring of force constant $K$ is cut away. The force constant of the remaining spring will be

A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are ${K_1}$ and ${K_2}$. The time period of the suspended mass will be

Two masses $m_1$ and $m_2$ connected by a spring of spring constant $k$ rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is

A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collies elastically with a rigid wall. After this collision :

$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.

$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.

$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.

$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.