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