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.

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 $\frac{d^{2}y}{d{t}^2}+Ky=0$ 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) $5\mathrm{\pi } \sec$

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:

A particle in SHM is described by the displacement equation $\mathrm{x}\left(\mathrm{t}\right)=\mathrm{Acos} \left(\mathrm{\omega t}+\mathrm{\theta }\right).$ If the initial position of the particle is 1 cm and its initial velocity is $\mathrm{\pi }$cm/s, what is its amplitude? (The angular frequency of the particle is $\pi s^{-1}$)

(1)   1 cm       

(2)  $\sqrt{2}$ cm

(3)    2 cm     

(4)   2.5 cm

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, then the time period becomes T₂. It can be concluded that: 

If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period will become:

1. 3T

2. 3/2T

3. 4T

4. 2T