Two springs of force constants $300 \, N/m$ (Spring $A$) and $400 \, N/m$ (Spring $B$) are joined together in series. The combination is compressed by $8.75 \, cm$. The ratio of energy stored in $A$ and $B$ is $\frac{E_A}{E_B}$. Then $\frac{E_A}{E_B}$ is equal to

$A$ mass $0.9 \, kg$,attached to a horizontal spring,executes $SHM$ with an amplitude $A_{1}$. When this mass passes through its mean position,a smaller mass of $124 \, g$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$,then the value of $\alpha$ will be $......$

$A$ load of mass $m$ falls from a height $h$ onto a scale pan hung from a spring as shown in the figure. If the spring constant is $k$,the mass of the scale pan is zero,and the mass $m$ does not bounce relative to the pan,then the amplitude of vibration is

$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.

$A$ 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

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