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

If a body is executing simple harmonic motion with frequency 'n', then the frequency of its potential energy is:

A spring is stretched by \(5~\text{cm}\) by a force \(10~\text{N}\). The time period of the oscillations when a mass of \(2~\text{kg}\) is suspended by it is:
1. \(3.14~\text{s}\)
2. \(0.628~\text{s}\)
3. \(0.0628~\text{s}\)
4. \(6.28~\text{s}\)

The radius of the circle, the period of revolution, initial position and direction of revolution are indicated in the figure.

The \(y\)-projection of the radius vector of rotating particle \(P\) will be:
1. \(y(t)=3 \cos \left(\frac{\pi \mathrm{t}}{2}\right)\), where \(y\) in m
2. \(y(t)=-3 \cos 2 \pi t\) , where \(y\) in m
3. \(y(t)=4 \sin \left(\frac{\pi t}{2}\right)\), where \(y\) in m
4. \(y(t)=3 \cos \left(\frac{3 \pi \mathrm{t}}{2}\right) \),  where \(y\) in m

The average velocity of a particle executing SHM in one complete vibration is:
1. zero
2. $\frac{A\omega }{2}$
3. $A\omega$
4. $\frac{A{\omega }^2}{2}$

The displacement of a particle executing simple harmonic motion is given by,

$y=A_0+A$ $\sin \omega t$ $+$ $B$ $\cos \omega t.$
Then the amplitude of its oscillation is given by:
1. $A+B$
2. $A_0$ $+$ $\sqrt{A^2+B^2}$
3. $\sqrt{A^2+B^2}$
4. $\sqrt{A_0^2}$ $+$ $(A+B{)}^2$

During simple harmonic motion of a body, the energy at the extreme position is: