The uniform stick of mass m length L is pivoted at the centre. In the equilibrium position shown in the figure, the identical light springs have their natural length. If the stick is turned through a small angle $\theta$, it executes SHM. The frequency of the motion is:

(1) $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{6K}{m}}$

(2) $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{3K}{2m}}$

(3) $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{3K}{m}}$

(4) None of these

Two simple pendulums have time periods T and $\frac{5T}{4}$. They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when bigger pendulum completes one oscillation will be:

1. $\frac{\mathrm{\pi }}{6}$

2. $\frac{\mathrm{\pi }}{4}$

3. $\frac{\mathrm{\pi }}{3}$

4. $\frac{\mathrm{\pi }}{2}$

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is 

(1) 0.8 T

(2) 0.25 T

(3) 2 T

(4) 4 T

When two displacements represented by y₁=asin(ωt) and y₂=bcos(ωt) are superimposed,the motion is -

(1) not a simple harmonic

(2) simple harmonic with amplitude a/b

(3) simple harmonic with amplitude $\sqrt{a^{2 }+ b^2}$

(4) simple harmonic with amplitude (a+b)/2

A S.H.M. has amplitude ‘a’ and  time period T. The maximum velocity will be -

(1) $\frac{4a}{T}$           

(2) $\frac{2a}{T}$

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

(4) $\frac{2\mathrm{\pi a}}{T}$                       

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

(1)  1 : 2                 

(2)  2 : 1

(3)  2 : 3                 

(4)  3 : 2

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3}cm/sec$
$$ will be

(1) $2\sqrt{3}cm$             

(2) $\sqrt{3}cm$

(3) 1 cm                   

(4) 2 cm

The maximum velocity of a simple harmonic motion represented by $y=3 \sin \left(100t+\frac{\mathrm{\pi }}{6}\right)$ is given by

(1)   300       

(2)   $\frac{3\mathrm{\pi }}{6}$

(3)   100         

(4)   $\frac{\mathrm{\pi }}{6}$

The displacement equation of a particle is $x=3\sin 2t+4\cos 2t$ The amplitude and maximum velocity will be respectively

(a)       5, 10       (b)         3, 2

c)         4, 2        (d)         3, 4