If a particle in SHM has a time period of \(0.1\) s and an amplitude of \(6\) cm, then its maximum velocity will be:
1. \(120 \pi\)$\mathrm{}$ cm/s 
2. \(0.6 \pi\)$\mathrm{}$ cm/s 
3. \(\pi\)$\mathrm{}$ cm/s
4. \(6\) cm/s

If the potential energy U (in J) of a body executing SHM is given by U = 20 + 10 (${\sin }^2$100$\mathrm{\pi }$t), then the minimum potential energy of the body will be:

The kinetic energy (K) of a simple harmonic oscillator varies with displacement (x) as shown. The period of the oscillation will be: (mass of oscillator is 1 kg)

The equation of an SHM is given as y=3sinωt + 4cosωt where y is in centimeters. The amplitude of the SHM will be?

The time periods for the figures (a) and (b) are ${\mathrm{T}}_1$ $\mathrm{and}$ ${\mathrm{T}}_2$ respectively. If all surfaces shown below are smooth, then the ratio $\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}$ will be:
   

1.  1: $\sqrt{3}$

2.  1: 1

3.  2: 1

4.  $\sqrt{3}$: 2

A particle is attached to a vertical spring and pulled down a distance of 0.01 m below its mean position and released. If its initial acceleration is 0.16 $\mathrm{m}/{\mathrm{s}}^2$, then its time period in seconds will be:

1.  $\mathrm{\pi }$

2.  $\frac{\mathrm{\pi }}{2}$

3.  $\frac{\mathrm{\pi }}{4}$

4.  $2\mathrm{\pi }$

A particle is executing linear simple harmonic motion with an amplitude \(a\) and an angular frequency \(\omega\). Its average speed for its motion from extreme to mean position will be:
1. \(\frac{a\omega}{4}\)
2. \(\frac{a\omega}{2\pi}\)
3. \(\frac{2a\omega}{\pi}\)
4. \(\frac{a\omega}{\sqrt{3}\pi}\)

Two simple harmonic motions, ${\mathrm{y}}_1$ $=$ $\mathrm{a}$ $\sin$ $\mathrm{\omega t}$ and ${\mathrm{y}}_2$ $=$ $2\mathrm{a}$ $\sin \left(\mathrm{\omega t}\right)$ $+$ $\frac{2\mathrm{\pi }}{3}$  are superimposed on a particle of mass m. The maximum kinetic energy of the particle will be:

1.  $\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{a}}^2$

2.  $\frac{5{\mathrm{m\omega }}^2{\mathrm{a}}^2}{4}$

3.  $\frac{3}{2}{\mathrm{m\omega }}^2{\mathrm{a}}^2$

4.  Zero

All the surfaces are smooth and springs are ideal. If a block of mass \(m\) is given the velocity \(v_0\) in the right direction, then the time period of the block shown in the figure will be:

                       
1. \(\frac{12l}{v_0}\)
2. \(\frac{2l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
3. \(\frac{4l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
4. \( \frac{\pi}{2}\sqrt{\frac{m}{k}}\)

In a spring pendulum, in place of mass, a liquid is used. If liquid leaks out continuously, then the time period of the spring pendulum: