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 particle executing simple harmonic motion of amplitude \(5~\text{cm}\) has a maximum speed of \(31.4~\text{cm/s}.\) The frequency of its oscillation will be:
1. \(1~\text{Hz}\)
2. \(3~\text{Hz}\)
3. \(2~\text{Hz}\)
4. \(4~\text{Hz}\)

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 circular motion of a particle with constant speed is:

Two springs of spring constants \(k_1\) and \(k_2\) are joined in series. The effective spring constant of the combination is given by:
1. \(\frac{k_1+k_2}{2}\)
2. \(k_1+k_2\)
3. \(\frac{k_1k_2}{k_1+k_2}\)
4. \(\sqrt{k_1k_2}{}\)

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