The amplitude of a simple harmonic oscillator is \(A\) and speed at the mean position is \(v_0\). The speed of the oscillator at the position \(x={A \over \sqrt{3}}\) will be:

1.	\(2v_0 \over \sqrt{3}\)	2.	\(\sqrt{2}v_0 \over 3\)
3.	\({2 \over 3}v_0\)	4.	\(\sqrt{\frac{2}{3}}v_0\)

A spring is having a spring constant k. It is cut into two parts A and B whose lengths are in the ratio of m:1. The spring constant of part A will be

1. $\frac{k}{m}$

2. $\frac{k}{m+1}$

3. k

4. $\frac{k(m+1)}{m}$

In damped oscillation, mass is 2 kg and spring constant is 500 N/m and damping coefficient is 1 kg s^–1. If the mass is displaced by 20 cm from its mean position and released, then what will be the value of its mechanical energy after 4 seconds?

1. 2.37 J

2. 1.37 J

3. 10 J

4. 5 J

1. \(x= 10\sin\left(\pi t+\frac{\pi}{6}\right)\)
2. \(x= 10\sin\left(\pi t\right)\)
3. \(x= 10\cos\left(\pi t\right)\)
4. \(x= 5\sin\left(\pi t+\frac{\pi}{6}\right)\)

All the surfaces are smooth and the system, given below, is oscillating with an amplitude \(\mathrm{A}.\) What is the extension of spring having spring constant \(\mathrm{k_1},\) when the block is at the extreme position?
             

Acceleration of the particle at \(t = \frac{8}{3}~\text{s}\) from the given displacement \((y)\) versus time \((t)\) graph will be?
                 
1. \(\frac{\sqrt{3}\pi^2}{4}~\text{cm/s}^2\)
2. \(-\frac{\sqrt{3}\pi^2}{4}~\text{cm/s}^2\)
3. \(-\pi^2~\text{cm/s}^2\)
4. zero

A body executes oscillations under the effect of a small damping force. If the amplitude of the body reduces by 50% in 6 minutes, then amplitude after the next 12 minutes will be [initial amplitude is ${\mathrm{A}}_0$] -

1.  $\frac{{\mathrm{A}}_0}{4}$

2.  $\frac{{\mathrm{A}}_0}{8}$

3.  $\frac{{\mathrm{A}}_0}{16}$

4.  $\frac{{\mathrm{A}}_0}{6}$

In a forced oscillation, when the system oscillates under the action of the driving force $\mathrm{F}$ $=$ ${\mathrm{F}}_0$ $\sin$ $\mathrm{\omega t}$ in addition to its internal restoring force, the particle oscillates with a frequency equal to

1.  The natural frequency of the body

2.  Frequency of driving force

3.  The difference in frequency of driving force and natural frequency

4.  Mean of the driving frequency and natural frequency

A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 ${\mathrm{t}}^2$, where y is in meters and t is in seconds. If g = 10 $\mathrm{m}/{\mathrm{s}}^2$, then the time period of the pendulum will be: