The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by :

(1) $X=a<mspace\; width="1em"></mspace>\sin 2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\times \mathrm{t}$

(2) $X=a<mspace\; width="1em"></mspace>\cos 2\mathrm{\pi }\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}$

(3) $X=a<mspace\; width="1em"></mspace>\sin \sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}$

(4) $X=a<mspace\; width="1em"></mspace>\cos \sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}$

A particle in SHM is described by the displacement function $x\left(t\right)=A<mspace\; width="1em"></mspace>\cos (\omega t+\varphi ),<mspace\; width="1em"></mspace>\omega =2\pi /T.$ If the initial (t = 0) position of the particle is 1 cm, its initial velocity is $\pi <mspace\; width="1em"></mspace>cm<mspace\; width="1em"></mspace>s^{-1}$ and its angular frequency is $\pi s^{-1}$, then the amplitude of its motion is:

(1) $\pi <mspace\; width="1em"></mspace>cm$

(2) 2 cm

(3) $\sqrt{2}<mspace\; width="1em"></mspace>cm$

(4) 1 cm

A particle starts SHM from the mean position. Its amplitude is 'a' and total energy E. At one instant its kinetic energy is 3E/4. Its displacement at this instant is :

(1) $y=a/\sqrt{2}$

(2) $y=\frac{a}{2}$

(3) $y=\frac{a}{\sqrt{3/2}}$

(4) y = a

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic?

(1) 1 cm

(2) $\sqrt{2}<mspace\; width="1em"></mspace>cm$

(3) 3 cm

(4) $2\sqrt{2}<mspace\; width="1em"></mspace>cm$

A body of mass 500 g is attached to a horizontal spring of spring constant 8 ${\mathrm{\pi }}^2{\mathrm{Nm}}^{-1}.$ If the body is pulled to a distance of 10 cm from its mean position, then its frequency of oscillation is :

(1) 2 Hz

(2) 4 Hz

(3) 8 Hz

(4) 0.5 Hz

Two springs of force constants k and 2k are connected to a mass m as shown below. The frequency of oscillation of the mass is:

(1) $\frac{1}{2\mathrm{\pi }}\sqrt{k/m}$

(2) $\frac{1}{2\mathrm{\pi }}\sqrt{2k/m}$

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

(4) $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{m}{k}}$

A weightless spring that has a force constant k oscillates with frequency n when a mass m is suspended from it. The spring is cut into equal halves and a mass 2m is suspended from one part of the spring. The frequency of oscillation will now become:

1. n

2. 2n

3. $\frac{\mathrm{n}}{\sqrt{2}}$

4. $\mathrm{n}{\left(2\right)}^{1/2}$

An object suspended from a spring exhibits oscillations of period T. Now, the spring is cut in two halves and the same object is suspended with two halves as shown in the figure. The new time period of oscillation will become :

1. $\frac{\mathrm{T}}{2\sqrt{2}}$

2. $\frac{T}{2}$

3. $\frac{T}{\sqrt{2}}$

4. 2T

A mass 1 kg suspended from a spring whose force constant is $400<mspace\; width="1em"></mspace>{\mathrm{Nm}}^{-1}$, executes simple harmonic oscillation. When the total energy of the oscillator is 2 J, the maximum acceleration experienced by the mass will be:

1. $2<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-2}$

2. $4<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-2}$

3. $40<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-2}$

4. $400<mspace\; width="1em"></mspace>{\mathrm{ms}}^{-2}$

What will be the force constant of the spring system shown in the figure?

1. $\frac{k_1}{2}+k_2$

2. ${\left[\frac{1}{2k_1}+\frac{1}{k_2}\right]}^{-1}$

3. $\frac{1}{2k_1}+\frac{1}{k_2}$

4. ${\left[\frac{2}{k_1}+\frac{1}{k_2}\right]}^{-1}$