If a pendulum giving correct time on the ground at a certain place is moved to the top of a tower 320 m high, then the loss in time (in seconds) measured by the pendulum clock in one week is:

1.  15.12

2.  7.72

3.  3.78

4.  30.24

Instantaneous acceleration (in ${\mathrm{ms}}^{-2}$) of a particle executing S.H.M. is given by $\mathrm{a} = -\frac{\mathrm{\pi }}{4}\sin \left(\frac{\mathrm{\pi }}{4}\mathrm{t} - \frac{\mathrm{\pi }}{6}\right)$.  The maximum speed of the particle will occur first time at

1.  1.75 s

2.  1.4 s

3.  1.2 s

4.  0.67 s

A particle moves according to the equation $\mathrm{x} = \mathrm{Acos}\frac{\mathrm{\pi }}{2}\mathrm{t}$. Distance covered by it in the time interval of t =0 to t =3 s is: (symbols have their usual meanings) 

1.  A

2.  4A

3.  3A

4.  2A

A body of mass m hanging with the help of three springs, each of spring constant k as shown. If the mass is slightly displaced and released, then the system will oscillate with the time period

1.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{3\mathrm{k}}}$

2.  $2\mathrm{\pi }\sqrt{\frac{2\mathrm{m}}{3\mathrm{k}}}$

3.  $2\mathrm{\pi }\sqrt{\frac{3\mathrm{m}}{2\mathrm{k}}}$

4.  $2\mathrm{\pi }\sqrt{\frac{3\mathrm{m}}{\mathrm{k}}}$

A block of mass m attached to a spring of constant k oscillates on a smooth surface. The other end of the spring is fixed to a wall. When spring is at its natural length, the speed of the block is v. Displacement of the block from its mean position before coming to instantaneous rest is:

1.  $\sqrt{\frac{\mathrm{mv}}{\mathrm{k}}}$

2.  $\mathrm{v}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

3.  $\mathrm{m}\sqrt{\frac{\mathrm{v}}{\mathrm{k}}}$

4.  $\frac{1}{\mathrm{k}}\sqrt{\frac{\mathrm{m}}{\mathrm{v}}}$

A particle under SHM has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s? The time period of oscillation (in second) is

1.  $\frac{\mathrm{\pi }}{2}$

2.  2$\mathrm{\pi }$

3.  $\mathrm{\pi }$

4.  4$\mathrm{\pi }$

The x-t graph of a particle undergoing SHM is shown below. The acceleration of the particle at $\mathrm{t} = \frac{4}{3}$ is:

1.  $\frac{\sqrt{3}}{32}{\mathrm{\pi }}^3 \mathrm{cm}/{\mathrm{s}}^2$

2.  $-\frac{{\mathrm{\pi }}^2}{32} \mathrm{cm}/{\mathrm{s}}^2$

3.  $\frac{{\mathrm{\pi }}^2}{32} \mathrm{cm}/{\mathrm{s}}^2$

4.  $-\frac{\sqrt{3}}{32}{\mathrm{\pi }}^2 \mathrm{cm}/{\mathrm{s}}^2$

The position-time (y - t) graph of a particle executing S.H.M. is shown. The time period of the particle is 4 seconds. Equation of particle executing S.H.M. is

1.  $\mathrm{y} = 6 \sin \left(2\mathrm{\pi t} + \frac{\mathrm{\pi }}{6}\right)$

2.  $\mathrm{y} = 6 \sin \left(\frac{\mathrm{\pi }}{2}\mathrm{t} + \frac{\mathrm{\pi }}{3}\right)$

3.  $\mathrm{y} = 6 \sin \left(\frac{\mathrm{\pi }}{2}\mathrm{t} + \frac{\mathrm{\pi }}{6}\right)$

4.  $\mathrm{y} = 6 \sin \left(\frac{\mathrm{\pi }}{2}\mathrm{t} - \frac{\mathrm{\pi }}{6}\right)$

A particle starts executing SHM from an extreme position with time period T and amplitude A. The distance travelled by the particle in time $\frac{15\mathrm{T}}{18}$is:

1.  3.5 A

2.  2.5 A

3.  0.5 A

4.  1.5