$A$ force of $6.4 \,N$ stretches a vertical spring by $0.1 \,m$. If it were to oscillate with a period of $\frac{\pi}{4} \,s$, then the mass that is to be suspended from the spring is:

$A$ particle of mass $m$ is attached to one end of a massless 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 collides 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 through the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{m}{k}}$.

The variation of potential energy of a harmonic oscillator is as shown in the figure. The spring constant is

Two springs of force constants $2k$ and $k$ are connected to a mass $m$ as shown. The mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is

The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?

$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.