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)   $(a^2-x^2)/X^2$

A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy changes into potential energy, will be:

1.   f/2         

2.  f

3.   2 f        

4.  4 f

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

­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , then the period of the pendulum will be

(1) T

(2) $\frac{\mathrm{T}}{4}$

(3) $\frac{2\mathrm{T}}{\sqrt{5}}$

(4) $2\mathrm{T}\sqrt{5}$

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

(1)    $\propto x$                 

(2)    $\propto x^2$

(3)   Independent of x 

(4)    $\propto x^{1/2}$

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

(1) $\sqrt{2\mathrm{gl}(1-\mathrm{sin\theta })}$

(2) $\sqrt{2\mathrm{gl}(1+\mathrm{cos\theta })}$

(3) $\sqrt{2\mathrm{gl}(1-\mathrm{cos\theta })}$

(4) $\sqrt{2\mathrm{gl}(1+\mathrm{sin\theta })}$

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.

In a simple pendulum, the period of oscillation T is related to length of the pendulum l as:
1. $\frac{\mathrm{l}}{T}$= constant
2. $\frac{{\mathrm{l}}^2}{T}$= constant
3. $\frac{\mathrm{l}}{T^2}$= constant
4. $\frac{{\mathrm{l}}^2}{T^2}$= constant