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 $y=3<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>\left(100t+\frac{\mathrm{\pi }}{6}\right)$ 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 $x=3\sin <mspace\; width="1em"></mspace>2t+4\cos <mspace\; width="1em"></mspace>2t<mspace\; width="1em"></mspace>$ 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 $x=A<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left(\omega t+\frac{\mathrm{\pi }}{4}\right)$ . 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. $\frac{2\mathrm{\pi }}{\omega }$

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

1  $144{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$       
2  $144m/se{c}^2$

3. $\frac{144}{{\mathrm{\pi }}^2}m/se{c}^2$          
4. $288{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$   

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

1.  – A Kx           

2.  A cos (Kx)

3.  A exp (– Kx) 

4.  A Kx

What is the maximum acceleration of the particle doing the SHM $y=2\sin \left[\frac{\mathrm{\pi t}}{2}+\varphi \right]$ where 2 is in cm

1.  $\frac{\mathrm{\pi }}{2}cm/s^2$                
2.  $\frac{{\mathrm{\pi }}^2}{2}cm/s^2$
3.  $\frac{\mathrm{\pi }}{4}cm/s^2$             
4.  $\frac{\mathrm{\pi }}{4}cm/s^2$

A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

1.  $\pm A$          

2.  Zero

3.  $\pm \frac{A}{2}$         

4.  $\pm \frac{A}{\sqrt{2}}$

The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)

1. $U=\frac{K{X}^2}{2}$                 

2. $U=K{X}^2$   

3.  $U=K$                       

4. $U=KX$ $$
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