A mass ' $M$ ' attached to a horizontal spring executes S.H.M. of amplitude $A_1$. When the mass M passes through its mean position, then a smaller mass ' $m$ ' is placed over it and both of them move together with amplitude $\mathrm{A}_2$. The ratio $\left(\frac{A_1}{A_2}\right)$ is

At a place, the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping amplitude same then the total energy will be

A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is $4 \mathrm{~N} / \mathrm{m}$. If at any instant its velocity is $20 \mathrm{~cm} / \mathrm{s}$, the displacement at that instant is (Amplitude of S.H.M. is 0.4 m )

The ratio of the frequencies of two simple pendulums is $4: 3$ at the same place. The ratio of their respective lengths is