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 ……….

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 $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.

Two springs with negligible masses and force constant of $K_1 = 200\, Nm^{-1}$ and $K_2 = 160\, Nm^{-1}$ are attached to the block of mass $m = 10\, kg$ as shown in the figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time $t = 0,$ a sharp impulse of $50\, Ns$ is given to the block with a hammer.

A mass $m$ is attached to two springs of same force constant $K$, as shown in following four arrangements. If $T_1, T_2, T_3$ and $T_4$ respectively be the time periods of oscillation in the following arrangements, in which case time period is maximum?

Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$, and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ….. $N$ ( take $g = 10\, ms^{-2}$)