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.  $\frac{2}{3}E$

A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

1. P.E. is maximum when x = 0

2. K.E. is maximum when x = 0

3. T.E. is zero when x = 0

4. K.E. is maximum when x is maximum

The total energy of a particle, executing simple harmonic motion is

1. $\propto <mspace\; width="1em"></mspace>x$                 

2. $\propto <mspace\; width="1em"></mspace>x^2$

3. Independent of x 

4.$\propto <mspace\; width="1em"></mspace>x^{1/2}$

A body is executing Simple Harmonic Motion. At a displacement x its potential energy is $E_1$ and at a displacement y its potential energy is $E_2$. The potential energy E at displacement $\left(x+y\right)$ is 

1.  $E=\sqrt{E_1}+\sqrt{E_2}$   

2.  $\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$

3.   $E=E_1+E_2$           

4.  None of these.

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

(1) $\frac{\mathrm{\pi }}{5}sec$       

(2) $2\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$

(3) $20\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$   

(4) $5\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$