The motion of a particle varies with time according to the relation $y= a\sin\omega t+ a\cos \omega t$. Then:

1.	the motion is oscillatory but not SHM.
2.	the motion is SHM with an amplitude $a\sqrt{2}$.
3.	the motion is SHM with an amplitude $\sqrt{2}$.
4.	the motion is SHM with an amplitude $a$.

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