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) $2\mathrm{T}\sqrt{5}$

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

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

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth then its time period on the other planet will be

(1) $\sqrt{\mathrm{T}}$

(2) T

(3) ${\mathrm{T}}^{1/3}$

(4) $\sqrt{2}$T