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 $g = 10 \ m/s^2$.

Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal,the ratio of the amplitude of $M$ to that of $N$ is

In the situation as shown in the figure,the time period of vertical oscillation of the block for small displacements will be:

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 $s$)

$A$ body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The mass of the spring is negligible. When the mass $m$ is pulled down slightly and released,it oscillates with a time period of $3 \ s$. When the mass $m$ is increased by $1 \ kg$,the time period of oscillation becomes $5 \ s$. What is the value of $m$ in $kg$?

$A$ particle of mass $1 \, \text{kg}$ is hanging from a spring of force constant $100 \, \text{N/m}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal is $\frac{T}{x}$. The value of $x$ is ..... .