The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
1. \(\frac{3\pi}{2}\text{rad}\)
2. \(\frac{\pi}{2}\text{rad}\)
3. zero
4. \(\pi ~\text{rad}\)

From the given functions, identify the function which represents a periodic motion:

The distance covered by a particle undergoing SHM in one time period is: (amplitude = A)
1. zero
2. A
3. 2 A
4. 4 A

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie in a straight line perpendicular to the paths of the two particles. The phase difference is:

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

If the time of mean position from amplitude (extreme) position is 6 seconds, then the frequency of SHM will be:

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:

The frequency of a spring is \(n\) after suspending mass \(M.\) Now, after mass \(4M\) mass is suspended from the spring, the frequency will be: