Two pendulums suspended from the same point have lengths of $2$ m and $0.5$ m. If they are displaced slightly and released, then they will be in the same phase when the small pendulum has completed:
1. $2$ oscillations
2. $4$ oscillations
3. $3$ oscillations
4. $5$ oscillations

A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:

1.  $2 \pi \sqrt{\frac{\left(\right. M   +   m \left.\right) l}{Mg}}$

2. $2 \pi \sqrt{\frac{ml}{Mg}}$

3. $2 \pi \sqrt{L   /   g}$

4. $2 \pi \sqrt{\frac{Ml}{\left(\right. m   +   M \left.\right) g}}$

The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

Two spherical bobs of masses $M_A$ and $M_B$ are hung vertically from two strings of length $l_A$ and $l_B$ respectively. If they are executing SHM with frequency as per the relation $f_A=2f_B,$ Then:
1. $l_A = \frac{l_B}{4}$
2. $l_A= 4l_B$
3. $l_A= 2l_B~\&~M_A=2M_B$
4. $l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}$

The total energy of the particle performing SHM depends on: 
1. $k,$ $a,$ $m$
2. $k,$ $a$
3. $k,$ $a$, $x$
4. $k,$ $x$

When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is $t _1$ and $t_2$ respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes $t_0.$ The correct relation between $t_0,$ $t_1$ & $t_2$ is:

1. ${{\mathrm{t}}_0}^2={{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2$

2. ${{\mathrm{t}}_0}^{-2}={{\mathrm{t}}_1}^{-2}+{{\mathrm{t}}_2}^{-2}$

3. ${{\mathrm{t}}_0}^{-1}={{\mathrm{t}}_1}^{-1}+{{\mathrm{t}}_2}^{-1}$

4. ${\mathrm{t}}_0={\mathrm{t}}_1+{\mathrm{t}}_2$

The time period of a mass suspended from a spring is $T$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:
1. $\frac{T}{4}$
2. $T$
3. $\frac{T}{2}$
4. $2T$