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 $8\sqrt{3}cm/sec$
$$ 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

(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)     $288{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$   

The displacement of an oscillating particle varies with time (in seconds) according to the equation $y\left(cm\right)=\sin \frac{\mathrm{\pi }}{2}\left(\frac{t}{2}+\frac{1}{3}\right)$ .The maximum acceleration of the particle is approximately 

(1) $5.21 cm/s^2$           

(2) $3.62 cm/s^2$

(3) $1.81 cm/s^2$           

(4) $0.62 cm/s^2$

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)   $\pm \frac{A}{\sqrt{2}}$