Displacement versus time curve for a particle executing SHM is shown in the figure. Choose the correct statement/s.

1.	Phase of the oscillator is the same at t =0 s and t = 2 s.
2.	Phase of the oscillator is the same at t =2 s and t=6 s.
3.	Phase of the oscillator is the same at t = 1 s and t=7 s.
4.	Phase of the oscillator is the same at t=1 s and t=5 s.

1.	1, 2 and 4	2.	1 and 3
3.	2 and 4	4.	3 and 4

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

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

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

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

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

A particle of mass \(m\) and charge \(\text-q\) moves diametrically through a uniformly charged sphere of radius \(R\) with total charge \(Q\). The angular frequency of the particle's simple harmonic motion, if its amplitude \(<R\), is given by:
1. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR} }\)
2. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^2} }\)
3. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^3}}\)
4. \( \sqrt{\dfrac{m}{4 \pi \varepsilon_0 ~qQ} }\)

From the given functions, identify the function which represents a periodic motion:

Identify the correct definition:

If the time of mean position from amplitude (extreme) position is 6 seconds, then the frequency of SHM will be:

Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then: 
1. $l_A=\frac{l_B}{4}$

2. $l_A=4l_B$

3. $l_A=2l_B$ $\&$ $M_A=2M_B$

4. $l_A=\frac{l_B}{2}$ $\&$ $M_A=\frac{M_B}{2}$

A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:

1.  \(2 \pi \sqrt{\frac{\left(\right. M   +   m \left.\right) l}{Mg}}\)

2. \(2 \pi \sqrt{\frac{ml}{Mg}}\)

3. \(2 \pi \sqrt{L   /   g}\)

4. \(2 \pi \sqrt{\frac{Ml}{\left(\right. m   +   M \left.\right) g}}\)

The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:

1. ${{\mathrm{t}}_0}^2={{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2$

2. ${{\mathrm{t}}_0}^{-2}={{\mathrm{t}}_1}^{-2}+{{\mathrm{t}}_2}^{-2}$

3. ${{\mathrm{t}}_0}^{-1}={{\mathrm{t}}_1}^{-1}+{{\mathrm{t}}_2}^{-1}$

4. ${\mathrm{t}}_0={\mathrm{t}}_1+{\mathrm{t}}_2$