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 \(t = 0~\text{s}~\text{and}~t = 2~\text{s}\).
2.	Phase of the oscillator is the same at \(t = 2~\text{s}~\text{and}~t = 6~\text{s}\).
3.	Phase of the oscillator is the same at \(t = 1~\text{s}~\text{and}~t = 7~\text{s}\).
4.	Phase of the oscillator is the same at \(t = 1~\text{s}~\text{and}~t = 5~\text{s}\).

1.	\(1,2~\text{and}~4\)	2.	\(1~\text{and}~3\)
3.	\(2~\text{and}~4\)	4.	\(3~\text{and}~4\)

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 \(\mathrm{x\text-}\)projection of the radius vector of the rotating particle \(P\) will be:

1. \(x \left( t \right) = B\) \(\text{sin} \left(\dfrac{2 πt}{30}\right)\)

2. \(x \left( t \right) = B\) \(\text{cos} \left(\dfrac{πt}{15}\right)\)

3. \(x \left( t \right) = B\) \(\text{sin} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)\)

4. \(x \left( t \right) = B\) \(\text{cos} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)\)

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. \( \sqrt{\dfrac{m}{4 \pi \varepsilon_0 ~qQ} }\)

Identify the correct definition:

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