$A$ block of mass $m$ attached to one end of a vertical spring produces an extension $x$. If the block is pulled and released,the periodic time of oscillation is:

$A$ mass $M$,attached to a horizontal spring,executes $S.H.M.$ with amplitude $A_1$. When the mass $M$ passes through its mean position,a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{A_1}{A_2}$ is

If the period of oscillation of mass $m$ suspended from a spring is $2 \ s$,then the period of oscillation of suspended mass $4m$ with the same spring will be: (in $s$)

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.

Two identical springs of constant $K$ are connected in series and parallel as shown in the figure. $A$ mass $M$ is suspended from them. The ratio of their frequencies in series to parallel combination will be

In the arrangement given in the figure,if the block of mass $m$ is displaced,the frequency is given by: