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)   $\pm \frac{A}{\sqrt{2}}$

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)   $(a^2-x^2)/X^2$

There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,$F=-Kx$ where x is the displacement. The total energy of body depends upon -

(1)   K, x         

(2)  K, a

(3)   K, a, x    

(4)  K, a, v

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

(1)    $\frac{1}{8}E$       

(2)        $\frac{1}{4}E$

(3)    $\frac{1}{2}E$       

(4)        $\frac{2}{3}E$