­­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

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.

The equation of motion of a particle is $\frac{d^{2}y}{d{t}^2}+Ky=0$ where K is positive constant. The time period of the motion is given by

(1) $\frac{2\mathrm{\pi }}{K}$             

(2) $2\mathrm{\pi K}$

(3) $\frac{2\mathrm{\pi }}{\sqrt{K}}$           

(4)  $2\pi \sqrt{K}$

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) $2\mathrm{\pi } \sec$

(3) $20\mathrm{\pi } \sec$   

(4) $5\mathrm{\pi } \sec$