Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

(1)   1 : 2                 

(2)  2 : 1

(3)   2 : 3                 

(4)  3 : 2

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

(a) $A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$        (b)  $A_1{{\omega }_1}^2=A_2{{\omega }_2}^2=A_3{A_3}^2$

(b) ${A_1}^2{\omega }_1={A_2}^2{\omega }_2={A_3}^2{\omega }_3$   (d) ${A_1}^2{{\omega }_1}^2={A_2}^2{{\omega }_2}^2=A^2$  

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

(a) 5, 10      

(b) 3, 2

(c) 4, 2       

(d) 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

(a)      $144{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$        (b)     $144m/se{c}^2$

c)      $\frac{144}{{\mathrm{\pi }}^2}m/se{c}^2$           (d)     $288{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$   

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

(a)  $\frac{\mathrm{\pi }}{2}cm/s^2$                 (b)    $\frac{{\mathrm{\pi }}^2}{2}cm/s^2$

(c)    $\frac{\mathrm{\pi }}{4}cm/s^2$               (d)    $\frac{\mathrm{\pi }}{4}cm/s^2$

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)   $\pm \frac{A}{\sqrt{2}}$