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. \(\sqrt{2} T\)

A body performs simple harmonic motion about x=0 with an amplitude a and a time period T. The speed of the body at $\mathrm{x}=\frac{\mathrm{a}}{2}$ will be:
1. $\frac{\mathrm{\pi a}\sqrt{3}}{2\mathrm{T}}$                                         
2. $\frac{\mathrm{\pi a}}{\mathrm{T}}$
3. $\frac{3{\mathrm{\pi }}^2\mathrm{a}}{\mathrm{T}}$                                           
4. $\frac{\mathrm{\pi a}\sqrt{3}}{\mathrm{T}}$

Two simple harmonic motions of angular frequency 100 rad s ^-1 and 1000 rad $s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. 1:10
2. 1:10²
3. 1:10³                               
4. 1:10⁴

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) $\sqrt{{\mathrm{g}}^2+{\mathrm{a}}^2}$