A particle executes linear simple harmonic motion with an amplitude of of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity  is equal to that of its acceleration. Then, its time period in seconds is 

(a) $\sqrt{\frac{5}{\pi }}$

(b)$\sqrt{\frac{5}{2\mathrm{\pi }}}$

(c)$\frac{4\mathrm{\pi }}{\sqrt{5}}$

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

A body mass m is attached to the lower end of a spring whose upper end is fixed. The spring has neglible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. The value of m in kg is-

(1) $\frac{3}{4}$               

(2)$\frac{4}{3}$

(3) $\frac{16}{9}$               

(4) $\frac{9}{16}$

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

(1)$kg m{s}^{-1}$                               

(2)$kg m{s}^{-2}$

(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. \(\sqrt{2} T\)

A spring of force constant \(k\) is cut into lengths of ratio \(1:2:3\). They are connected in series and the new force constant is \(k'\). Then they are connected in parallel and the force constant is \(k''\). Then \(k':k''\) is:
1. \(1:9\)
2. \(1:11\)
3. \(1:14\)
4. \(1:6\)

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is:

A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then its time period of vibration will be: