Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

When a body of mass $1.0 \,kg$ is suspended from a certain light spring hanging vertically, its length increases by $5 \,cm$. By suspending a $2.0 \,kg$ block to the spring and if the block is pulled through $10 \,cm$ and released, the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 \,m/s^2$)

$A$ spring-mass system is oscillating horizontally. What will be the effect on the time period if the spring is made to oscillate vertically?

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_{1}$ and $K_{2}$ respectively. If the maximum velocities during oscillations are equal,the ratio of the amplitude of $A$ and $B$ is

$A$ light spring is suspended with mass $m_1$ at its lower end and its upper end fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ seconds. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate,the period is found to be $\frac{3}{2} T$. The ratio $m_1 : m_2$ is

An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially,the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is: