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

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

(1) $\frac{\mathrm{\pi }}{5}sec$       

(2) $2\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$

(3) $20\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$   

(4) $5\mathrm{\pi }<mspace\; width="1em"></mspace>\sec$

A simple harmonic wave having an amplitude a and time period T is represented by the equation $y=5<mspace\; width="1em"></mspace>\mathrm{sin\pi }\left(t+4\right)$m Then the value of amplitude (a) in (m) and time period  (T) in second are       

1.   $a=10,<mspace\; width="1em"></mspace>T=2$   

2. $a=5,<mspace\; width="1em"></mspace>T=1$

3.    $a=10,<mspace\; width="1em"></mspace>T=1$    

4. $a=5,<mspace\; width="1em"></mspace>T=2$

If the displacement equation of a particle be represented by $y=A\sin Pt+<mspace\; width="1em"></mspace>B\cos <mspace\; width="1em"></mspace>Pt$ , the particle executes

1.         A uniform circular motion

2.         A uniform elliptical motion

3.         A S.H.M.

4         A rectilinear motion

On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K, the period of oscillation of the body (assuming the springs as massless) is

1. $2\mathrm{\pi }{\left(\frac{M}{2K}\right)}^{1/2}$         
2. $2\mathrm{\pi }{\left(\frac{2\mathrm{M}}{K}\right)}^{1/2}$
3. $2\mathrm{\pi }\frac{\mathrm{Mg}<mspace\; width="1em"></mspace>\mathrm{sin\theta }}{2\mathrm{K}}$         
4. $2\mathrm{\pi }{\left(\frac{2\mathrm{Mg}}{\mathrm{K}}\right)}^{1/2}$

The graph shows the variation of displacement of a particle executing SHM with time. We infer from this graph that:

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

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

For a particle executing simple harmonic motion, the kinetic energy K is given by $k=k_{o<mspace\; width="1em"></mspace>}{\cos }^2<mspace\; width="1em"></mspace>\omega t$. The maximum value of potential energy is

1. $K_o$                     

2. Zero

3. $\frac{K_o}{2}$                 

4. Not obtainable