A body at the end of a spring executes S.H.M. with a period ${\mathrm{T}}_1$ while the corresponding period for another spring is ${\mathrm{T}}_2$. If the period of oscillation with two springs in series is T, then:

1.  $\mathrm{T} = {\mathrm{T}}_1 + {\mathrm{T}}_2$

2.  ${\mathrm{T}}^2 = {\mathrm{T}}_1^2 + {\mathrm{T}}_2^2$

3.  $\frac{1}{\mathrm{T}} = \frac{1}{{\mathrm{T}}_1} + \frac{1}{{\mathrm{T}}_2}$

4.  $\frac{1}{{\mathrm{T}}^2} = \frac{1}{{\mathrm{T}}_1^2} + \frac{1}{{\mathrm{T}}_2^2}$

A body is executing linear S.H.M. At a position x, its potential energy is ${\mathrm{E}}_1$, and at a position y, its potential energy is ${\mathrm{E}}_2$. The potential energy at the position (x + y) is

1.  ${\mathrm{E}}_1 + {\mathrm{E}}_2$

2.  $\sqrt{{\mathrm{E}}_1^2 + {\mathrm{E}}_2^2}$

3.  ${\mathrm{E}}_1 + {\mathrm{E}}_2 + 2\sqrt{{\mathrm{E}}_1{\mathrm{E}}_2}$

4.  $\sqrt{{\mathrm{E}}_1{\mathrm{E}}_2}$

The equation of a particle executing simple harmonic motion is $y = 0.4 \sin \left(2\mathrm{\pi t} + \frac{\mathrm{\pi }}{3}\right)$ (where t is in seconds and y is in meters). The initial phase of the particle is:

1.  $2\mathrm{\pi t} + \frac{\mathrm{\pi }}{3}$

2.  $\frac{\mathrm{\pi }}{3}$

3.  $\frac{7\mathrm{\pi }}{3}$

4.  $\frac{8\mathrm{\pi }}{3}$

An ideal spring-mass system has a time period of vibration T. If the spring is cut into 4 identical parts and same mass oscillates with one of these parts, then the new time period of vibration will be

1.  $\frac{\mathrm{T}}{2}$

2.  T

3.  $\frac{\mathrm{T}}{4}$

4.  2T

The equation of a particle executing simple harmonic motion is $y=2\sqrt{2}\sin 314t.$ Displacement y from the mean position where acceleration becomes zero is: (y is in cm and t is in second) 

1.  2 cm

2.  0

3.  $\frac{1}{\sqrt{2}}$ cm

4.  $2\sqrt{2}$ cm

The displacement \((\mathrm{x})\) of an SHM varies with time \((\mathrm{t})\) as shown in the figure. The frequency of variation of potential energy is:

A simple pendulum bob is a hollow sphere full of sand suspended by means of a wire. If all the sand is drained out immediately, then the time period of the pendulum will:

A simple pendulum of length L is suspended from the ceiling of a cart which is sliding without friction on an inclined plane of inclination $\mathrm{\theta }$. The time period of the pendulum is

1.  $\mathrm{T} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

2.  $\mathrm{T} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g} \mathrm{cos\theta }}}$

3.  $\mathrm{T} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g} \mathrm{sin\theta }}}$

4.  $\mathrm{T} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g} \mathrm{tan\theta }}}$

A body is placed on a horizontal platform that is undergoing vertical SHM. If the amplitude of oscillation is 40 cm, then the least period of oscillation for which an object placed over the platform is not detached from it is:

1.  1.256 s

2.  12.56 s

3.  0.1256 s

4.  125.6 s

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