$A$ block $A$ of mass $1\, kg$ is connected to two identical springs of spring constant $800\, N/m$ and is placed on a smooth horizontal surface as shown in the figure. Initially,the springs are relaxed. Now,the block $A$ is slightly displaced to the left and released. The time period of oscillation of the system is

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction is

$A$ $1 \, kg$ block attached to a spring vibrates with a frequency of $1 \, Hz$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8 \, kg$ block placed on the same table. The frequency of vibration of the $8 \, kg$ block is ..... $Hz$.

$A$ mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_1$ and $K_2$. For the frictionless surface,the time period of oscillation of mass $m$ is

$A$ spring stretches by $2 \text{ mm}$ when it is loaded with a mass of $200 \text{ g}$. From the equilibrium position, the mass is further pulled down by $2 \text{ mm}$ and released. The frequency associated with the system and the maximum energy in the spring are . . . . . . $\text{Hz}$ and . . . . . . $\text{J}$, respectively. (Take $g = 10 \text{ m/s}^2$)

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.