A partricle 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 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 displacement of a particle moving in S.H.M. at any instant is given by $y=a \sin$$\omega t$ . The acceleration after time $t=\frac{T}{4}$ (where T is the time period) -

1. $a\omega$                         

2.$-a\omega$

3. $a{\omega }^2$                         

4.  $-a{\omega }^2$

A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

1.  $\pm A$          

2.  Zero

3.  $\pm \frac{A}{2}$         

4.  $\pm \frac{A}{\sqrt{2}}$

The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)

1. $U=\frac{K{X}^2}{2}$                 

2. $U=K{X}^2$   

3.  $U=K$                       

4. $U=KX$ $$
$$

­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , then the period of the pendulum will be

1. T

2. $\frac{\mathrm{T}}{4}$

3. $\frac{2\mathrm{T}}{\sqrt{5}}$

4. $2\mathrm{T}\sqrt{5}$

The total energy of a particle, executing simple harmonic motion is

1. $\propto x$                 

2. $\propto x^2$

3. Independent of x 

4.$\propto x^{1/2}$

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}\text{'}}}$,  where $g^{\text{'}}$   is equal to

1. g                                                       

2. g-a

3. g+a

4. $\sqrt{{\mathrm{g}}^2+{\mathrm{a}}^2}$