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

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $\mathrm{t}=\frac{\mathrm{T}}{6}$, the ratio of the potential energy to kinetic energy of the particle is

$$ \left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right] $$

Two particles ' $A$ ' and ' $B$ ' execute SHMs of periods ' $T$ ' and $\frac{3 T}{2}$. If they start from the mean position then the phase difference between them, when the particle ' $A$ ' completes two oscillations will be