An SHM has an amplitude \(a\) and a time period \(T.\) The maximum velocity will be:
1. \({4a \over T}\)       
2. \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)

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. $\frac{2\mathrm{\pi }}{\omega }$

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

1.  – A Kx           

2.  A cos (Kx)

3.  A exp (– Kx) 

4.  A Kx

For a particle executing simple harmonic motion, the kinetic energy K is given by $k=k_{o }{\cos }^2 \omega t$. The maximum value of potential energy is

1. $K_o$                     

2. Zero

3. $\frac{K_o}{2}$                 

4. Not obtainable

There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,$F=-Kx$ where x is the displacement. The total energy of body depends upon -

1.  K, x         

2.  K, a

3.  K, a, x    

4.  K, a, v

A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

1. P.E. is maximum when x = 0

2. K.E. is maximum when x = 0

3. T.E. is zero when x = 0

4. K.E. is maximum when x is maximum

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