A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

(1) $\frac{1}{2}s$

(2) $\frac{1}{6}s$

(3) $\frac{1}{4}s$

(4) $\frac{1}{3}s$

The uniform stick of mass m length \(\text L\) is pivoted at the centre. In the equilibrium position shown in the figure, the identical light springs have their natural length. If the stick is turned through a small angle $\theta$, it executes SHM. The frequency of the motion is:

1. \(\frac{1}{2 \pi} \sqrt{\frac{6 K}{m}} \)

2. \(\frac{1}{2 \pi} \sqrt{\frac{3 K}{2 m}} \)

3. \(\frac{1}{2 \pi} \sqrt{\frac{3 K}{m}} \)

4. None of these

Two simple pendulums have time periods T and $\frac{5T}{4}$. They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when bigger pendulum completes one oscillation will be:

1. $\frac{\mathrm{\pi }}{6}$

2. $\frac{\mathrm{\pi }}{4}$

3. $\frac{\mathrm{\pi }}{3}$

4. $\frac{\mathrm{\pi }}{2}$

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector \(\vec a\) is correctly shown in: 

1.		2.
3.		4.

A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket:

(1) Comes down with uniform acceleration

(2) Moves around the earth in a geostationary orbit

(3) Moves up with a uniform velocity

(4) Moves up with the uniform acceleration