Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has  completed how many oscillations?

1. 5

2. 1

3. 2

4. 3

The displacement of a particle is represented by the equation $\mathrm{y}=3 \cos \left(\frac{\mathrm{\pi }}{4}-2\mathrm{\omega t}\right)$. The motion of the particle is.

1. simple harmonic with period $2\pi /\mathrm{\omega }$

2. simple harmonic with period $\mathrm{\pi }/\mathrm{\omega }$

3. periodic but not simple harmonic

4. non-periodic

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is the maximum speed?

1. 100 m/min

2. 200 m/min

3. 300 m/min

4. 50 m/min

Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillations of two points?

1. $\mathrm{\pi }$

2. $\frac{\mathrm{\pi }}{6}$

3. $\frac{\mathrm{\pi }}{3}$

4. $\frac{2\mathrm{\pi }}{3}$

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) $\frac{1}{2}$