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. $\frac{2\mathrm{\pi }}{\omega }$

The displacement of a particle moving in S.H.M. at any instant is given by $y=a \sin$$\omega t$ . The acceleration after time $t=\frac{T}{4}$ (where T is the time period) -

1. $a\omega$                         

2.$-a\omega$

3. $a{\omega }^2$                         

4.  $-a{\omega }^2$

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. $U=KX$ $$
$$

The angular velocity and the amplitude of a simple pendulum is $\omega$ and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is 

1,  $X^2{\omega }^2\left(a^2-X^2{\omega }^2\right)$       

2.  $X^2/\left(a^2-x^2\right)$

3.  $\left(a^2-X^2{\omega }^2\right)/X^2{\omega }^2$       

4. $(a^2-x^2)/X^2$