Two simple harmonic motions of angular frequency $100~\text{rad s}^{-1}$ and $1000~\text{rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. $1:10$
2. $1:10^{2}$
3. $1:10^{3}$
4. $1:10^{4}$

A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin $\left(\omega t +\pi /6\right)$.After the elapse of what fraction of the time period the velocity of the point will be equal to half to its maximum velocity?

1.  $\frac{T}{8}$

2.  $\frac{T}{6}$

3. $\frac{\mathrm{ T}}{3}$

4 $\frac{T}{\mathrm{ 12}}$

The angular velocities of three bodies in simple harmonic motion are ${\omega }_1,{\omega }_2,{\omega }_3$ with their respective amplitudes as $A_1,A_2,A_3$. If all the three bodies have same mass and maximum velocity, then

1. $A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$       
2.  $A_1{{\omega }_1}^2=A_2{{\omega }_2}^2=A_3{A_3}^2$
3. ${A_1}^2{\omega }_1={A_2}^2{\omega }_2={A_3}^2{\omega }_3$   
4. ${A_1}^2{{\omega }_1}^2={A_2}^2{{\omega }_2}^2=A^2$  

The maximum velocity of a simple harmonic motion represented by $y=3 \sin \left(100t+\frac{\mathrm{\pi }}{6}\right)$ is given by

1. 300       

2.  $\frac{3\mathrm{\pi }}{6}$

3. 100         

4.  $\frac{\mathrm{\pi }}{6}$

The displacement equation of a particle is $x=3\sin 2t+4\cos 2t$ The amplitude and maximum velocity will be respectively

1. 5, 10      

2. 3, 2

3. 4, 2       

4. 3, 4