A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a string of length L. Time period of pendulum is T₀. When parallel plates are charged, the electric field between the plates is E and time period changes to T. The ratio T/T₀ is equal to 

(1) ${\left(\frac{{\displaystyle g+\frac{{\displaystyle qE}}{{\displaystyle m}}}}{{\displaystyle g}}\right)}^{1/2}$           (2) ${\left(\frac{{\displaystyle g}}{{\displaystyle g+\frac{{\displaystyle qE}}{{\displaystyle m}}}}\right)}^{3/2}$

(3) ${\left(\frac{{\displaystyle g}}{{\displaystyle g+\frac{{\displaystyle qE}}{{\displaystyle m}}}}\right)}^{1/2}$           (4) None of these 

A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is  β. Then, its time period of vibration will be

1. β²/α² 
 

2. α/β
 

3. β²/α
 

4. 2πβ/α 

The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are 

1.$\mathrm{ k}g m{s}^{-1}$                               

2.$\mathrm{ k}g m{s}^{-2}$

3. $kg s^{-1}$                                   

4. $kg s$

The period of oscillation of a mass \(M\) suspended from a spring of negligible mass is \(T.\) If along with it another mass \(M\) is also suspended, the period of oscillation will now be:
1. \(T\)
2. \(T/\sqrt{2}\)
3. \(2T\)
4. \(\sqrt{2} T\)

Two simple harmonic motions of angular frequency \(100~\text{rad s}^{-1}\) and \(1000~\text{rad s}^{-1}\) have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. \(1:10\)
2. \(1:10^{2}\)
3. \(1:10^{3}\)
4. \(1:10^{4}\)

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 $8\sqrt{3}\mathrm{cm}/\sec$ 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 instantaneous displacement of a simple pendulum oscillator is given by $x=A \cos \left(\omega t+\frac{\mathrm{\pi }}{4}\right)$ . Its speed will be maximum at time

1. $\frac{\mathrm{\pi }}{4\omega }$      

2. $\frac{\mathrm{\pi }}{2\omega }$

3. $\frac{\mathrm{\pi }}{\omega }$                

4. $\frac{2\mathrm{\pi }}{\omega }$