Which one of the following equations of motion represents simple harmonic motion? (where \(k,k_0,k_1~\text{and}~a\) are all positive.)
1. Acceleration \(=-k_0x+k_1x^2\)
2. Acceleration \(=-k(x+a)\)
3. Acceleration \(=k(x+a)\)
4. Acceleration \(=kx\)

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 amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3}\mathrm{cm}/\sec$ will be

1.   $2\sqrt{3}cm$             

2.  $\sqrt{3}cm$

3.  1 cm                   

4.  2 cm

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

The instantaneous displacement of a simple pendulum oscillator is given by $x=A \cos \left(\omega t+\frac{\mathrm{\pi }}{4}\right)$ . Its speed will be maximum at time

1. $\frac{\mathrm{\pi }}{4\omega }$      

2. $\frac{\mathrm{\pi }}{2\omega }$

3. $\frac{\mathrm{\pi }}{\omega }$                

4. $\frac{2\mathrm{\pi }}{\omega }$