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

Two simple pendulums have first (A) bob of mass ' $M_1$ ' and length ' $L_1$ ', second (B) of mass ' $\mathrm{M}_2$ ' and length ' $\mathrm{L}_2$ '. $\mathrm{M}_1=\mathrm{M}_2$ and $\mathrm{L}_1=2 \mathrm{~L}_2$. If their total energies are same then the correct statement is

As shown in the figure, $S_1$ and $S_2$ are identical springs with spring constant K each. The oscillation frequency of the mass ' $m$ ' is ' $f$ '. If the spring $\mathrm{S}_2$ is removed, the oscillation frequency will become