A body has a time period ${\mathrm{T}}_1$ under the action of one force and ${\mathrm{T}}_2$ under the action of another force, the square of the time period when both the forces are acting in the same direction is

1. ${\mathrm{T}}_1^2{\mathrm{T}}_2^2$

2. ${\mathrm{T}}_1^2{\mathrm{T}}_2^2$

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

4. ${\mathrm{T}}_1^2{\mathrm{T}}_2^2/({\mathrm{T}}_1^2+{\mathrm{T}}_2^2)$

Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is:

(1) x(t) = B sin $\left(\frac{2\mathrm{\pi t}}{30}\right)$

(2) x(t) = B cos $\left(\frac{\mathrm{\pi t}}{15}\right)$

(3) x(t) = B sin $\left(\frac{\mathrm{\pi t}}{15}+\frac{\mathrm{\pi }}{2}\right)$

(4) x(t) = B $\left(\frac{\mathrm{\pi t}}{15}+\frac{\mathrm{\pi }}{2}\right)$

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and the ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3 $and b_{air} =1.3$, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to (ln 5 = 1.601, ln 2 = 0.693) :

(1) 231 s

(2) 208 s

(3) 161 s

(4) 142 s

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; x = $a_1$cos$\omega t$, and y = $a_2\cos 2\omega t$ traces a curve given by:

(1)

(2)

(3)

(4)

The angular frequency of the damped oscillator is given by, $\omega =\sqrt{\frac{k}{m}-\frac{r^2}{4m^2}}$where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $r^2/mk$ is 8%, the change in the time period compared to the undamped oscillator is approximately as follows:

(1) increases by 1%

(2) increases by 8%

(3) decreases by 1%

(4) decreases by 8%

A pendulum with a time period of 1s is losing energy due to damping. At a certain time, its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant $(in s^{-1})$ is:

(1) $\frac{1}{30} \ln 3$

(2) $\frac{1}{15} \ln 3$

(3) 2

(4) $\frac{1}{2}$

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are the same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement -A/2 and they are moving towards each other. If they cross each other at time t, then t is:

(1) T/6

(2) 5T/6

(3) T/3

(4) T/4

A block of mass 0.1 kg is connected to an elastic spring constant 640 ${\mathrm{Nm}}^{-1}$ and oscillates in a damping medium of damping constant ${10}^{-2} \mathrm{kg} {\mathrm{s}}^{-1}$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibrations to drop to half of its initial value is closest to:

1. 2s

2. 3.5 s

3. 5 s

4. 7 s

 A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half of its value for every 10 oscillations. The time it will take to drop to $\frac{1}{1000}$ of the original amplitude is close to:

1. 100 s

2. 20 s

3. 10 s

4. 50 s

A simple pendulum oscillating in air has time period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}th$ of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:

1, $4\mathrm{T}\sqrt{\frac{1}{15}}$

2. $2\mathrm{T}\sqrt{\frac{1}{10}}$

3. $4\mathrm{T}\sqrt{\frac{1}{14}}$

4. $2\mathrm{T}\sqrt{\frac{1}{14}}$