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

­­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.$\propto x^{1/2}$

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}\text{'}}}$,  where $g^{\text{'}}$   is equal to

1. g                                                       

2. g-a

3. g+a

4. $\sqrt{{\mathrm{g}}^2+{\mathrm{a}}^2}$

If the length of second's pendulum is decreased by 2%, how many seconds it will lose per day?

1. 3927 sec

2. 3727 sec

3. 3427 sec

4. 864 sec