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 has 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 the part \(A\) will be:
1. \(\dfrac{k}{m}\)
2. \(\dfrac{k}{m+1}\)
3. \(k\)
4. \(\dfrac{k(m+1)}{m}\)

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 \({A}.\) What is the extension of spring having spring constant \({k_1},\) 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 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:

The graph between the velocity \((v)\) of a particle executing SHM and its displacement \((x)\) is shown in the figure. The time period of oscillation for this SHM will be:

      
1. \(\sqrt{\frac{\alpha}{\beta}}\)
2. \(2\pi\sqrt{\frac{\alpha}{\beta}}\)
3. \(2\pi\left(\frac{\beta}{\alpha}\right)\)
4. \(2\pi\left(\frac{\alpha}{\beta}\right)\)