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. \(\pi ~\text{rad}\)

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. $\sqrt{A_0^2}$ $+$ $(A+B{)}^2$

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

A particle executes linear simple harmonic motion with amplitude of \(3~\text{cm}\). When the particle is at \(2~\text{cm}\) from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
1. \(\frac{\sqrt5}{2\pi}\)
2. \(\frac{4\pi}{\sqrt5}\)
3. \(\frac{4\pi}{\sqrt3}\)
4. \(\frac{\sqrt5}{\pi}\)

A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then its time period of vibration will be:

When two displacements are represented by \(y_1 = a \text{sin}(\omega t)\) and \(y_2 = b\text{cos}(\omega t)\) are superimposed, then the motion is:

A particle is executing SHM along a straight line. Its velocities at distances \(x_1\) and \(x_2\) from the mean position are \(v_1\) and \(v_2\), respectively. Its time period is:

The oscillation of a body on a smooth horizontal surface is represented by the equation, \(X=A \text{cos}(\omega t)\),
where \(X=\) displacement at time \(t,\) \(\omega=\) frequency of oscillation.
Which one of the following graphs correctly shows the variation of acceleration, \(a\) with time, \(t?\)
(\(T=\) time period) \(a~~O~~T~~t~~\)

1.		2.
3.		4.