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

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:

A spring pendulum is on the rotating table. The initial angular velocity of the table is \(\omega_{0}\) and the time period of the pendulum is \(T_{0}.\) Now the angular velocity of the table becomes \(2\omega_{0},\) then the new time period will be:
1. \(2T_{0}\)
2. \(T_0\sqrt{2}\)
3. remains the same
4. \(\frac{T_0}{\sqrt{2}}\)

Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then: 
1. $l_A=\frac{l_B}{4}$

2. $l_A=4l_B$

3. $l_A=2l_B$ $\&$ $M_A=2M_B$

4. $l_A=\frac{l_B}{2}$ $\&$ $M_A=\frac{M_B}{2}$

The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

A simple pendulum is pushed slightly from its equilibrium towards left and then set free to execute simple harmonic motion. Select the correct graph between its velocity(\(v\)) and displacement (\(x \)).

1.		2.
3.		4.

The period of oscillation of a simple pendulum of length \(\mathrm{L}\) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\theta$, is given by:

1.   $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gcos\theta }}}$               

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gsin}\theta }}$

3.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g} }}$                     

4. $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gtan\theta }}}$

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: 

Two simple pendulums of length 1 m and 16 m are in the same phase at the mean position at any instant. If T is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:

1.  $\frac{3\mathrm{T}}{2}$

2.  $\frac{3\mathrm{T}}{4}$

3.  $\frac{2\mathrm{T}}{3}$

4.  $\frac{4\mathrm{T}}{3}$