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