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

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

The equation of motion of a particle is \({d^2y \over dt^2}+Ky=0 \) where \(K\) is a positive constant. The time period of the motion is given by: 

The kinetic energy of a particle executing SHM 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 will be:
1. $\frac{\mathrm{\pi }}{5}sec$ 
2. $2\mathrm{\pi }$ $\sec$
3. $20\mathrm{\pi }$ $\sec$   
4. $5\mathrm{\pi }$ $\sec$

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