A body oscillates with SHM according to the equation (in SI units), \(x= 5\cos\left[2\pi t +\frac{\pi}{4}\right].\) At \(t = 1.5\) s, acceleration of the body will be:

1.	\(140 \text{ cm} / \text{s}^2 \)	2.	\(160 \text{ m} / \text{s}^2 \)
3.	\(140 \text{ m} / \text{s}^2 \)	4.	\(14 \text{ m} / \text{s}^2\)

The potential energy of a simple harmonic oscillator, when the particle is halfway to its endpoint, will be:
1. \(\frac{2E}{3}\)
2. \(\frac{E}{8}\)
3. \(\frac{E}{4}\)
4. \(\frac{E}{2}\)

A particle of mass \(m\) oscillates with simple harmonic motion between points \(x_1\) and \(x_2\), the equilibrium position being \(O\). Its potential energy is plotted. It will be as given below in the graph:

1.		2.
3.		4.

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

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

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

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. \(l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}\)