The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to one-third,then the same mass oscillates with frequency $v_2$. Then:

$A$ mass '$m_1$' is suspended from a spring of negligible mass. The spring is pulled slightly in the downward direction and released; the mass performs $S.H.M.$ with a period '$T_1$'. If the mass is increased by '$m_2$',the time period becomes '$T_2$'. The ratio $\frac{m_2}{m_1}$ is

Two blocks of masses $m$ and $M$ $(M > m)$ are placed on a frictionless table as shown in the figure. $A$ massless spring with spring constant $k$ is attached to the lower block. If the system is slightly displaced and released,then ($\mu =$ coefficient of friction between the two blocks):
$(A)$ The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{M + m}{k}}$
$(B)$ The acceleration of the blocks is $a = \frac{kx}{M + m}$ ($x =$ displacement of the blocks from the mean position)
$(C)$ The magnitude of the frictional force on the upper block is $f = \frac{mkx}{M + m}$
$(D)$ The maximum amplitude of the upper block,if it does not slip,is $A = \frac{\mu mg(M + m)}{mk} = \frac{\mu g(M + m)}{k}$ (Wait,let's re-evaluate: $f_{max} = \mu mg$. Since $f = ma = m \cdot \frac{kx}{M+m}$,at max amplitude $A$,$m \cdot \frac{kA}{M+m} = \mu mg \implies A = \frac{\mu g(M+m)}{k}$)
$(E)$ Maximum frictional force can be $\mu mg$.
Choose the correct answer from the options given below.

$A$ 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 time period of oscillation for the given combination will be:

How does the period of oscillation depend on the mass of the block attached to the end of a spring?