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. \(4~\text{Hz}\)

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

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

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

The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is:

The average energy in one time period in simple harmonic motion is:
1. \(\dfrac{1}{2} m \omega^{2} A^{2}\)
2. \(\dfrac{1}{4} m \omega^{2} A^{2}\)
3. \(m \omega^{2} A^{2}\)
4. zero

A spring-mass system oscillates with a frequency \(\nu.\) If it is taken in an elevator slowly accelerating upward, the frequency will:
1. increase
2. decrease
3. remain same
4. become zero

A particle moves in a circular path with a continuously increasing speed. Its motion is:
1. periodic
2. oscillatory
3. simple harmonic
4. none of them