A particle executes linear SHM between \(x=A.\) The time taken to go from \(0\) to \(A/2\) is \(T_1\) and to go from \(A/2\) to \(A\) is \(T_2\) then:

1.	\(T_1<T_2\)	2.	\(T_1>T_2\)
3.	\(T_1=T_2\)	4.	\(T_1= 2T_2\)

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. \(2\pi\left(\frac{\alpha}{\beta}\right)\)

Acceleration-time (\(a\text-t\)) graph for a particle performing SHM is shown in the figure. Select the incorrect statement.

A simple pendulum is pushed slightly from its equilibrium towards the left and then set free to execute the simple harmonic motion. Select the correct graph between its velocity (\(v\)) and displacement (\(x \)).

1.		2.
3.		4.

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 spring-block system oscillates with a time period \(T\) on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes \(T'.\) Then, one can conclude that:
1. \(T'>T\)
2. \(T'<T\)
3. \(T'=T\)
4. \(T'=2T\)